Chapter 10 · Additional Maths · Trigonometric identities, equations and proofs

Trigonometric identities, equations and proofs: formulae, solving and the R form

A handful of identities turn a hard trigonometric expression into an easy one. This chapter uses the Pythagorean and compound-angle identities, the double and half angle formulae and the R form a cos t + b sin t = R cos(t - alpha) to find exact values, prove identities, solve equations in a given interval and model tides and geometry. Every equation here is solved in a stated interval, and only the angles inside it are listed.

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Chapter
10

Figure 1 · a cosine wave: amplitude 2 (the R of a cos t + b sin t is the amplitude of the combined wave)

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Compound and double angles

sin(A ± B) = sin A cos B ± cos A sin B, cos(A ± B) = cos A cos B ∓ sin A sin B. Set B = A to get the double-angle formulae; pick the form of cos 2A that uses the ratio you were given.

Solve in the interval, never in general

Find the basic angle, use the quadrant to list the angles, then add whole turns until the range of the inner angle is covered. Factorise: never divide by sin x or cos x unless you have shown it cannot be 0 at a solution.

R form: one wave from two

$a\cos\theta + b\sin\theta = R\cos(\theta - \alpha)$ with $R = \sqrt{a^2 + b^2}$ and $\tan\alpha = \dfrac{b}{a}$ . Greatest value $R$ , least $-R$ .

By the end of this chapter

Use sin²A + cos²A = 1, sec²A = 1 + tan²A and cosec²A = 1 + cot²A, and find every ratio of an angle from one ratio and its quadrant.
Expand sin(A ± B), cos(A ± B) and tan(A ± B), and find the exact value of a non-special angle such as 75° or 105°.
Use sin 2A, the three forms of cos 2A and tan 2A, and the half-angle results cos²(x/2) = (1 + cos x)/2 and sin²(x/2) = (1 - cos x)/2.
Prove simple identities from the left side to the right side, and use a proved identity to solve an equation (the hence strategy).
Solve trigonometric equations in a given interval, including scaled and shifted angles, quadratics in one ratio, factorised equations and radians.
Write a cos t ± b sin t in the form R cos(t ± α) or R sin(t ± α), find its greatest and least values, solve equations with it and use it in models.

✶

Start here

Maths In Real Life

Three situations where an angle or a wave decides a fit, a depth or an area. The pictures come first; the working in each solution comes from the same numbers.

$Front view of an empty workshop doorway with plain straight jambs, a flat floor and a plain wall. A rectangular plywood sheet is drawn tilted across the open doorway so that its two outer corners just touch the two jambs; the angle of its long edge to the floor and the three lengths are labelled$

120\text{ cm}

35\text{ cm}

110\text{ cm}

\theta

Maths In Real Life 1

A plywood sheet through a workshop doorway

A rectangular sheet turned through an angle needs a doorway at least as wide as its horizontal shadow, and that width is a sum of a cosine and a sine: exactly the shape of a cosine wave with a larger amplitude.

In a school workshop a rectangular plywood sheet $120$ cm long and $35$ cm wide is slid through a doorway that is $110$ cm wide, in the plane of the doorway. The long edge of the sheet makes an angle $\theta$ with the floor, $0^{\circ} \le \theta \le 90^{\circ}$ , and the sheet needs a width of $(120\cos\theta + 35\sin\theta)$ cm across the doorway.

(a)Write $(120\cos\theta + 35\sin\theta)$ in the form $R\cos(\theta - \alpha)$ , where $R > 0$ and $\alpha$ is acute, and give $R$ and $\alpha$ . Hence find the greatest width the sheet can need and the angle $\theta_0$ at which this happens.
(b)Find the angle $\theta$ at which the width needed is exactly $110$ cm, correct to $1$ decimal place.

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Colour keybasic angle or value found on the wayfinal value

(a)

Expanding:

R\cos\left(\theta - \alpha\right) = R\cos\theta\cos\alpha + R\sin\theta\sin\alpha

Expand

R\cos(\theta - \alpha)

with the compound-angle formula, then match it with

a\cos\theta + b\sin\theta

: the coefficients of

\cos\theta

and

\sin\theta

must agree.

Matching coefficients gives

R\cos\alpha = 120

and

R\sin\alpha = 35

\displaystyle R\;= \sqrt{120^2 + 35^2} = \textcolor{#010105}{125}

Square both equations and add:

\cos^2\alpha + \sin^2\alpha = 1

leaves

R^2 = a^2 + b^2

R

is the positive root.

\displaystyle \tan\alpha\;= \dfrac{R\sin\alpha}{R\cos\alpha} = \dfrac{35}{120} = \dfrac{7}{24}

Divide the two matching equations:

R

cancels. Both

R\cos\alpha

and

R\sin\alpha

are positive, so

\alpha

is acute.

\displaystyle \alpha\;= \tan^{-1}\left(\dfrac{7}{24}\right) \approx \textcolor{#010105}{16.3^{\circ}}

Both

R\cos\alpha

and

R\sin\alpha

are positive, so

\alpha

is acute: Use the calculator in degree mode and give 1 decimal place.

\alpha

was found in the working above.

\displaystyle \text{greatest value}\;= 125

The cos of any angle lies between

-1

and

1

, so

R\cos(\ldots)

lies between

-R

and

R

: the greatest value is

R

The greatest value is reached when

\cos\left(\theta_0 - \alpha\right) = 1

The maximum of cos is at the angle where it equals

1

\displaystyle \theta_0 - \alpha\;= 0

The angle whose cos is

1

in one turn.

\displaystyle \theta_0\;= 0 + 16.3^{\circ} = \textcolor{#010105}{16.3^{\circ}}

The angle is

\alpha

itself, or its negative when the form is the other way round.

(b)

Expanding:

R\cos\left(\theta - \alpha\right) = R\cos\theta\cos\alpha + R\sin\theta\sin\alpha

Expand

R\cos(\theta - \alpha)

with the compound-angle formula, then match it with

a\cos\theta + b\sin\theta

: the coefficients of

\cos\theta

and

\sin\theta

must agree.

Matching coefficients gives

R\cos\alpha = 120

and

R\sin\alpha = 35

\displaystyle R\;= \sqrt{120^2 + 35^2} = \textcolor{#010105}{125}

Square both equations and add:

\cos^2\alpha + \sin^2\alpha = 1

leaves

R^2 = a^2 + b^2

R

is the positive root.

\displaystyle \tan\alpha\;= \dfrac{R\sin\alpha}{R\cos\alpha} = \dfrac{35}{120} = \dfrac{7}{24}

Divide the two matching equations:

R

cancels. Both

R\cos\alpha

and

R\sin\alpha

are positive, so

\alpha

is acute.

\displaystyle \alpha\;= \tan^{-1}\left(\dfrac{7}{24}\right) \approx \textcolor{#010105}{16.260^{\circ}}

Both

R\cos\alpha

and

R\sin\alpha

are positive, so

\alpha

is acute: Use the calculator in degree mode and give 1 decimal place.

Using the R form found above (with

\alpha \approx 16.260^{\circ}

120\cos\theta + 35\sin\theta \approx 125\cos\left(\theta - 16.260^{\circ}\right)

The R form found above.

\displaystyle 125\cos\left(\theta - 16.260^{\circ}\right)\;= 110

Write the equation in the R form.

\displaystyle \cos\left(\theta - 16.260^{\circ}\right)\;= \dfrac{110}{125} = \dfrac{22}{25}

Divide by

R

Let

\phi = \theta - 16.26^{\circ}

. Since

0^{\circ} \le \theta \le 90^{\circ}

, the angle

\phi

runs from

-16.26^{\circ}

73.74^{\circ}

, so list every angle with this ratio in that range.

\displaystyle \text{smallest angle}\;= 0^{\circ} - 16.26^{\circ} = \textcolor{#010105}{-16.26^{\circ}}

Smallest angle: put the smallest

\theta

into

\phi

\displaystyle \text{largest angle}\;= 90^{\circ} - 16.26^{\circ} = \textcolor{#010105}{73.74^{\circ}}

Largest angle: put the largest

\theta

into

\phi

\displaystyle \beta\;= \cos^{-1}\left(\dfrac{22}{25}\right) \approx \textcolor{#010105}{28.358^{\circ}}

The basic angle

\beta

is the acute angle with this ratio: use the inverse function on the positive value.

\displaystyle \phi_1\;= \textcolor{#010105}{28.358^{\circ}}

Since cosine is positive in quadrants I and IV, the angles with this ratio in one full turn are:

\displaystyle \phi_2\;= 360^{\circ} - 28.358^{\circ} = \textcolor{#010105}{331.642^{\circ}}

The other angle in the same turn, from the symmetry of the quadrants.

\phi_2 = 331.642^{\circ}

is outside the range

-16.26^{\circ} \le \phi \le 73.74^{\circ}

, so it is not used as it stands.

\displaystyle \theta_1\;= 28.358^{\circ} + 16.26^{\circ} \approx \textcolor{#010106}{44.6^{\circ}}

Add

16.26^{\circ}

back to return from

\phi

\theta

Answer

(a)

R = 125

;

\alpha = 16.3^{\circ}

; max value

= 125

;

\theta_0 = 16.3^{\circ}

(b)

\theta = 44.6^{\circ}

$A side view of the end of a small wooden fishing jetty on calm water, with round wooden piles standing in the water and one white mooring post on the deck.$

h

3\text{ m}

\text{side view}

\text{seabed}

Maths In Real Life 2

The depth of water at a jetty

Near the coast the water level rises and falls twice a day with the tide, so a harbour master has to know when the water is deep enough for a boat to dock. A cosine model with a time scale in degrees per hour is exactly what a tide table is built from.

The depth of water, $h$ metres, at the end of a jetty $t$ hours after midnight is modelled by $h = 2.5 + 0.6\cos 30t + 0.8\sin 30t$ , where the angle is measured in degrees. A boat can dock only when the depth is at least $3$ m. Consider the first 12 hours after midnight.

(a)Express $h$ in the form $c + R\cos(kt - \alpha)$ , where $R > 0$ and $\alpha$ is acute.
(b)Find the greatest depth and the least depth of the water.
(c)Find the times when the depth is exactly $3$ m. Give your answers correct to 2 decimal places.
(d)Find the intervals of time during which the boat can dock.

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Colour keybasic angle or value found on the wayfinal value

(a)

Let

\theta = 30t

, so that

h = 2.5 + \dfrac{3}{5}\cos\theta + \dfrac{4}{5}\sin\theta

with the angle

\theta

in degrees.

Expanding:

R\cos\left(\theta - \alpha\right) = R\cos\theta\cos\alpha + R\sin\theta\sin\alpha

Expand

R\cos(\theta - \alpha)

with the compound-angle formula, then match it with

a\cos\theta + b\sin\theta

: the coefficients of

\cos\theta

and

\sin\theta

must agree.

Matching coefficients gives

R\cos\alpha = \dfrac{3}{5}

and

R\sin\alpha = \dfrac{4}{5}

\displaystyle R\;= \sqrt{\dfrac{9}{25} + \dfrac{16}{25}} = \textcolor{#010105}{1}

Square both equations and add:

\cos^2\alpha + \sin^2\alpha = 1

leaves

R^2 = a^2 + b^2

R

is the positive root.

\displaystyle \tan\alpha\;= \dfrac{R\sin\alpha}{R\cos\alpha} = \dfrac{\dfrac{4}{5}}{\dfrac{3}{5}} = \dfrac{4}{3}

Divide the two matching equations:

R

cancels. Both

R\cos\alpha

and

R\sin\alpha

are positive, so

\alpha

is acute.

\displaystyle \alpha\;= \tan^{-1}\left(\dfrac{4}{3}\right) \approx \textcolor{#010105}{53.1^{\circ}}

Both

R\cos\alpha

and

R\sin\alpha

are positive, so

\alpha

is acute: Use the calculator in degree mode and give 1 decimal place.

Putting

R = 1

and

\alpha \approx 53.1^{\circ}

back, and

\theta = 30t

again:

h \approx 2.5 + \cos\left(30t - 53.1^{\circ}\right)

(angle to 1 decimal place).

Put

R

and

\alpha

back into the form.

(b)

\displaystyle h_{\text{max}}\;= 2.5 + 1 = \textcolor{#010106}{3.5}

The cosine lies between

-1

and

1

, so

R\cos(\ldots)

lies between

-R

and

R

: the greatest value of

h

is the constant plus

R

\displaystyle h_{\text{min}}\;= 2.5 - 1 = \textcolor{#010106}{1.5}

The cosine lies between

-1

and

1

, so

R\cos(\ldots)

lies between

-R

and

R

: the least value of

h

is the constant minus

R

(c)

\displaystyle 2.5 + \cos\left(\theta - 53.1^{\circ}\right)\;= 3

Put

h

equal to the level, with

\theta = 30t

for the angle inside the brackets.

\displaystyle \cos\left(\theta - 53.1^{\circ}\right)\;= \dfrac{3 - \left(\dfrac{5}{2}\right)}{1} = \dfrac{1}{2}

Subtract the constant from both sides, then divide by

R

Let

\phi = \theta - 53.1^{\circ}

. Since

0^{\circ} \le \theta \le 360^{\circ}

, the angle

\phi

runs from

-53.1^{\circ}

306.9^{\circ}

, so list every angle with this ratio in that range.

\displaystyle \text{smallest angle}\;= 0^{\circ} - 53.1^{\circ} = \textcolor{#010105}{-53.1^{\circ}}

Smallest angle: put the smallest

\theta

into

\phi

\displaystyle \text{largest angle}\;= 360^{\circ} - 53.1^{\circ} = \textcolor{#010105}{306.9^{\circ}}

Largest angle: put the largest

\theta

into

\phi

\displaystyle \beta\;= \cos^{-1}\left(\dfrac{1}{2}\right) = \textcolor{#010105}{60^{\circ}}

The basic angle

\beta

is the acute angle with this ratio: use the inverse function on the positive value.

\displaystyle \phi_1\;= \textcolor{#010105}{60^{\circ}}

Since cosine is positive in quadrants I and IV, the angles with this ratio in one full turn are:

\displaystyle \phi_2\;= 360^{\circ} - 60^{\circ} = \textcolor{#010105}{300^{\circ}}

The other angle in the same turn, from the symmetry of the quadrants.

\displaystyle \theta_1\;= 60^{\circ} + 53.1^{\circ} = \textcolor{#010106}{113.1^{\circ}}

Add

53.1^{\circ}

back to return from

\phi

\theta

\displaystyle \theta_2\;= 300^{\circ} + 53.1^{\circ} = \textcolor{#010106}{353.1^{\circ}}

Add

53.1^{\circ}

back to return from

\phi

\theta

\displaystyle t\;= \dfrac{113.1^{\circ}}{30^{\circ}} \approx \textcolor{#010106}{3.77}

Divide the angle by the number of degrees per unit of

t

. Give 2 decimal places.

\displaystyle t\;= \dfrac{353.1^{\circ}}{30^{\circ}} \approx \textcolor{#010106}{11.77}

Divide the angle by the number of degrees per unit of

t

. Give 2 decimal places.

(d)

h

is continuous, so between two consecutive times it cannot cross the level: it is above the level for the whole gap or below it for the whole gap. Test one value of

t

half way along each gap, including the gaps at the ends of the interval.

\displaystyle t_{\text{test}}\;= \dfrac{0 + 3.77}{2} = 1.885

Half way between

0

and

3.77

\displaystyle h(1.885)\;= 2.5 + \dfrac{3}{5}\cos 56.55^{\circ} + \dfrac{4}{5}\sin 56.55^{\circ} \approx 3.5

Calculator in degree mode, 2 decimal places: this is above the level

3

, so

h

is above it for the whole gap.

\displaystyle t_{\text{test}}\;= \dfrac{3.77 + 11.77}{2} = 7.77

Half way between

3.77

and

11.77

\displaystyle h(7.77)\;= 2.5 + \dfrac{3}{5}\cos 233.1^{\circ} + \dfrac{4}{5}\sin 233.1^{\circ} \approx 1.5

Calculator in degree mode, 2 decimal places: this is below the level

3

, so

h

is below it for the whole gap.

\displaystyle t_{\text{test}}\;= \dfrac{11.77 + 12}{2} = 11.885

Half way between

11.77

and

12

\displaystyle h(11.885)\;= 2.5 + \dfrac{3}{5}\cos 356.55^{\circ} + \dfrac{4}{5}\sin 356.55^{\circ} \approx 3.05

Calculator in degree mode, 2 decimal places: this is above the level

3

, so

h

is above it for the whole gap.

Joining the gaps where

h

is above the level, and keeping the end values (where

h

equals the level), gives the answer.

Answer

(a)

h \approx 2.5 + \cos\left(30t - 53.1^{\circ}\right)

(b) greatest value of

h

= 3.5

; least value of

h

= 1.5

(c)

t = 3.77, 11.77

(d)

h \ge 3

for

0 \le t \le 3.77 \text{ and } 11.77 \le t \le 12

$Plan view of a flat open green field with a thin grey path round its edge. A triangular plot is outlined on the grass: one fence runs along the bottom, another rises to the upper right from the left corner, and a third closes the triangle; the angle between the first two is marked$

80\text{ m}

60\text{ m}

\theta

1200\text{ m}^{2}

\text{plan view}

Maths In Real Life 3

A triangular plot of grass

Two fences and the angle between them fix the size of a triangular plot: the area is half the product of the fences and the sine of the angle between them, and one value of the area is shared by an acute and an obtuse angle.

A triangular plot of grass is bounded by two straight fences of length $80$ m and $60$ m that meet at the angle $\theta$ . The plot has an area of $1200$ square m, and a third fence closes the triangle.

(a)Find every possible value of the angle $\theta$ between the two fences.
(b)Find the length of the third fence when $\theta$ is the acute angle found in (a), correct to $1$ decimal place.
(c)Find the length of the third fence when $\theta$ is the obtuse angle found in (a), correct to $1$ decimal place.

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Colour keybasic angle or value found on the wayfinal value

(a)

\displaystyle \text{Area}\;= \dfrac{1}{2} \times 80 \times 60 \times \sin\theta = 2400\sin\theta

Area of a triangle from two sides and the angle between them:

\tfrac{1}{2}bc\sin A

, a formula on the formula page.

\displaystyle 2400\sin\theta\;= 1200

The area is given, so put the expression equal to it.

\displaystyle \sin\theta\;= \dfrac{1200}{2400} = \textcolor{#010105}{\dfrac{1}{2}}

Divide both sides by the coefficient of

\sin\theta

An angle of a triangle lies between

0^{\circ}

and

180^{\circ}

, so list every angle in that range with this sine: one acute and one obtuse.

\displaystyle \alpha\;= \sin^{-1}\left(\dfrac{1}{2}\right) = \textcolor{#010105}{30^{\circ}}

The basic angle

\alpha

is the acute angle with this ratio: use the inverse function on the positive value.

\displaystyle \theta_1\;= \textcolor{#010105}{30^{\circ}}

Since sine is positive in quadrants I and II, the angles with this ratio in one full turn are:

\displaystyle \theta_2\;= 180^{\circ} - 30^{\circ} = \textcolor{#010105}{150^{\circ}}

The other angle in the same turn, from the symmetry of the quadrants.

(b)

\displaystyle a^{2}\;= 80^{2} + 60^{2} - 2 \times 80 \times 60 \times \cos 30^{\circ}

Cosine rule:

a

is the side opposite the angle between the sides of length

80

and

60

\displaystyle = 10000 - 9600 \times \dfrac{\sqrt{3}}{2} = \textcolor{#010105}{10000 - 4800 \sqrt{3}}

Put in the exact value of the cosine and multiply out.

\displaystyle a \approx \textcolor{#010106}{41.1}

Take the positive square root of the right-hand side above with a calculator and round to 1 decimal place at the end.

(c)

\displaystyle a^{2}\;= 80^{2} + 60^{2} - 2 \times 80 \times 60 \times \cos 150^{\circ}

Cosine rule:

a

is the side opposite the angle between the sides of length

80

and

60

\displaystyle = 10000 - 9600 \times - \dfrac{\sqrt{3}}{2} = \textcolor{#010105}{4800 \sqrt{3} + 10000}

Put in the exact value of the cosine and multiply out.

\displaystyle a \approx \textcolor{#010106}{135.3}

Take the positive square root of the right-hand side above with a calculator and round to 1 decimal place at the end.

Answer

(a)

\theta = 30^{\circ} \text{ or } 150^{\circ}

(b) third side

\approx 41.1

\approx 135.3

Supports K341-G1-06

Identities from one ratio: Pythagoras and the quotient

One trigonometric ratio of an angle, together with the quadrant the angle is in, fixes every other ratio of that angle. The tool is the Pythagorean identity: the three ratios of one angle are not independent, and the quadrant decides every sign.

Revision. In a right-angled triangle

\sin\theta = \dfrac{\text{opp}}{\text{hyp}}

\cos\theta = \dfrac{\text{adj}}{\text{hyp}}

and

\tan\theta = \dfrac{\text{opp}}{\text{adj}}

, and

\text{opp}^{2} + \text{adj}^{2} = \text{hyp}^{2}

. The signs by quadrant (All, Sine, Tangent, Cosine) are in Chapter 9.

The Core Idea

The Pythagorean and quotient identities.

For every angle $A$ : $\sin^{2}A + \cos^{2}A = 1$ , $\sec^{2}A = 1 + \tan^{2}A$ and $\operatorname{cosec}^{2}A = 1 + \cot^{2}A$ . These three are on the formula page. The quotient identities $\tan A = \dfrac{\sin A}{\cos A}$ and $\cot A = \dfrac{\cos A}{\sin A}$ are not: memorise them.

SGSchoolKaki Tip

After you take a square root, state the quadrant sign first: in quadrant II, $\cos A$ is negative, so $\cos A$ is minus the square root of $1 - \sin^{2}A$ , never the positive root by habit.

The sign of each ratio in each quadrant (All, Sine, Tangent, Cosine)
Quadrant	sin	cos	tan	Picture
I	positive	positive	positive	All three ratios are positive
II	positive	negative	negative	Only sine is positive
III	negative	negative	positive	Only tangent is positive
IV	negative	positive	negative	Only cosine is positive

Example 1.1

One ratio and a quadrant: sine given

Given that $\sin A = \dfrac{5}{13}$ and that $A$ lies in the second quadrant, find the exact values of $\cos A$ and $\tan A$ .

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Colour keybasic angle or value found on the way

The basic-angle triangle has opposite 5 and hypotenuse 13.

\displaystyle \text{adj}\;= \sqrt{13^2 - 5^2} = \sqrt{144} = \textcolor{#010105}{12}

Pythagoras gives the third side of the triangle.

\displaystyle \cos A\;= -\dfrac{12}{13}

In quadrant II only sine (and cosec) is positive: cos is negative, so the answer is negative.

\displaystyle \tan A\;= -\dfrac{5}{12}

In quadrant II only sine (and cosec) is positive: tan is negative, so the answer is negative.

Answer

\cos A = - \dfrac{12}{13}

;

\tan A = - \dfrac{5}{12}

Your Turn 1.1

Given that $\sin A = - \dfrac{8}{17}$ and that $A$ lies in the third quadrant, find the exact values of $\cos A$ and $\tan A$ .

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Colour keybasic angle or value found on the wayfinal value

The basic-angle triangle has opposite 8 and hypotenuse 17.

\displaystyle \text{adj}\;= \sqrt{17^2 - 8^2} = \sqrt{225} = \textcolor{#010105}{15}

Pythagoras gives the third side of the triangle.

\displaystyle \cos A\;= -\dfrac{15}{17}

In quadrant III only tangent (and cot) is positive: cos is negative, so the answer is negative.

\displaystyle \tan A\;= \textcolor{#010106}{\dfrac{8}{15}}

In quadrant III only tangent (and cot) is positive: tan is positive, so the answer is positive.

Answer

\cos A = - \dfrac{15}{17}

;

\tan A = \dfrac{8}{15}

Example 1.2

Tangent given, reciprocal asked

Given that $\tan A = \dfrac{7}{24}$ and that $A$ lies in the third quadrant, find the exact values of $\sin A$ , $\cos A$ and $\operatorname{cosec} A$ .

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Colour keybasic angle or value found on the way

The basic-angle triangle has opposite 7 and adjacent 24.

\displaystyle \text{hyp}\;= \sqrt{7^2 + 24^2} = \sqrt{625} = \textcolor{#010105}{25}

Pythagoras gives the third side of the triangle.

\displaystyle \sin A\;= -\dfrac{7}{25}

In quadrant III only tangent (and cot) is positive: sin is negative, so the answer is negative.

\displaystyle \cos A\;= -\dfrac{24}{25}

In quadrant III only tangent (and cot) is positive: cos is negative, so the answer is negative.

\displaystyle \operatorname{cosec} A\;= -\dfrac{25}{7}

In quadrant III only tangent (and cot) is positive: cosec is negative, so the answer is negative.

Answer

\sin A = - \dfrac{7}{25}

;

\cos A = - \dfrac{24}{25}

;

\operatorname{cosec} A = - \dfrac{25}{7}

Your Turn 1.2

Given that $\tan A = - \dfrac{20}{21}$ and that $A$ lies in the fourth quadrant, find the exact values of $\sin A$ , $\cos A$ and $\operatorname{cosec} A$ .

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Colour keybasic angle or value found on the wayfinal value

The basic-angle triangle has opposite 20 and adjacent 21.

\displaystyle \text{hyp}\;= \sqrt{20^2 + 21^2} = \sqrt{841} = \textcolor{#010105}{29}

Pythagoras gives the third side of the triangle.

\displaystyle \sin A\;= -\dfrac{20}{29}

In quadrant IV only cosine (and sec) is positive: sin is negative, so the answer is negative.

\displaystyle \cos A\;= \textcolor{#010106}{\dfrac{21}{29}}

In quadrant IV only cosine (and sec) is positive: cos is positive, so the answer is positive.

\displaystyle \operatorname{cosec} A\;= -\dfrac{29}{20}

In quadrant IV only cosine (and sec) is positive: cosec is negative, so the answer is negative.

Answer

\sin A = - \dfrac{20}{29}

;

\cos A = \dfrac{21}{29}

;

\operatorname{cosec} A = - \dfrac{29}{20}

A square root has two signs.

\cos^{2}A = \dfrac{16}{25}

gives

\cos A = \dfrac{4}{5}

\cos A = -\dfrac{4}{5}

, and only the quadrant tells you which. Taking the positive root in every quadrant is the most common lost mark in this chapter.

Example 1.3

Cosine given, secant asked

Given that $\cos A = \dfrac{20}{29}$ and that $A$ lies in the fourth quadrant, find the exact values of $\sin A$ , $\tan A$ and $\sec A$ .

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Colour keybasic angle or value found on the wayfinal value

The basic-angle triangle has adjacent 20 and hypotenuse 29.

\displaystyle \text{opp}\;= \sqrt{29^2 - 20^2} = \sqrt{441} = \textcolor{#010105}{21}

Pythagoras gives the third side of the triangle.

\displaystyle \sin A\;= -\dfrac{21}{29}

In quadrant IV only cosine (and sec) is positive: sin is negative, so the answer is negative.

\displaystyle \tan A\;= -\dfrac{21}{20}

In quadrant IV only cosine (and sec) is positive: tan is negative, so the answer is negative.

\displaystyle \sec A\;= \textcolor{#010106}{\dfrac{29}{20}}

In quadrant IV only cosine (and sec) is positive: sec is positive, so the answer is positive.

Answer

\sin A = - \dfrac{21}{29}

;

\tan A = - \dfrac{21}{20}

;

\sec A = \dfrac{29}{20}

Your Turn 1.3

Given that $\cos A = - \dfrac{21}{29}$ and that $A$ lies in the second quadrant, find the exact values of $\sin A$ , $\tan A$ and $\sec A$ .

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Colour keybasic angle or value found on the wayfinal value

The basic-angle triangle has adjacent 21 and hypotenuse 29.

\displaystyle \text{opp}\;= \sqrt{29^2 - 21^2} = \sqrt{400} = \textcolor{#010105}{20}

Pythagoras gives the third side of the triangle.

\displaystyle \sin A\;= \textcolor{#010106}{\dfrac{20}{29}}

In quadrant II only sine (and cosec) is positive: sin is positive, so the answer is positive.

\displaystyle \tan A\;= -\dfrac{20}{21}

In quadrant II only sine (and cosec) is positive: tan is negative, so the answer is negative.

\displaystyle \sec A\;= -\dfrac{29}{21}

In quadrant II only sine (and cosec) is positive: sec is negative, so the answer is negative.

Answer

\sin A = \dfrac{20}{29}

;

\tan A = - \dfrac{20}{21}

;

\sec A = - \dfrac{29}{21}

Think About It

Why does

\sec^{2}A = 1 + \tan^{2}A

follow from

\sin^{2}A + \cos^{2}A = 1

? Try dividing every term of the first identity by

\cos^{2}A

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\displaystyle \dfrac{\sin^{2}A + \cos^{2}A}{\cos^{2}A}\;= \dfrac{1}{\cos^{2}A}

Divide the left side of

\sin^2 A + \cos^2 A = 1

\cos^2 A

; the right side

1

becomes

\dfrac{1}{\cos^2 A}

\displaystyle \dfrac{\sin^{2}A}{\cos^{2}A} + \dfrac{\cos^{2}A}{\cos^{2}A}\;= \tan^{2}A + 1

Split the fraction:

\dfrac{\sin A}{\cos A} = \tan A

and the last term is

1

Answer

Both left-hand sides are the same expression, so

\tan^{2}A + 1 = \dfrac{1}{\cos^{2}A} = \sec^{2}A

. The same division by

\sin^{2}A

gives

1 + \cot^{2}A = \operatorname{cosec}^{2}A

The same triangle idea evaluates a nested expression such as $\tan\left(\cos^{-1}\dfrac{3}{5}\right)$ . Name the inner angle $\theta$ , so $\cos\theta = \dfrac{3}{5}$ . The principal value of $\cos^{-1}$ lies between $0^{\circ}$ and $180^{\circ}$ and that of $\tan^{-1}$ lies between $-90^{\circ}$ and $90^{\circ}$ , which tells you the sign of the other ratios.

Example 1.4

Tangent of an inverse cosine

Using exact values only, with no calculator, evaluate $\tan\left(\cos^{-1}\left(\dfrac{3}{5}\right)\right)$ .

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Colour keybasic angle or value found on the wayfinal value

Let

\theta = \cos^{-1}\left(\dfrac{3}{5}\right)

\displaystyle \cos\theta\;= \dfrac{3}{5}

The principal value of

\cos^{-1}

is an angle on the upper half of the unit circle, so

\sin\theta \ge 0

\displaystyle \sin\theta\;= \sqrt{1 - \left(\dfrac{3}{5}\right)^2} = \sqrt{\dfrac{16}{25}} = \textcolor{#010105}{\dfrac{4}{5}}

Pythagorean identity, positive root because

\sin\theta \ge 0

on this range.

\displaystyle \tan\theta\;= \dfrac{\sin\theta}{\cos\theta} = \dfrac{\dfrac{4}{5}}{\dfrac{3}{5}} = \textcolor{#010106}{\dfrac{4}{3}}

Quotient identity.

Answer

\tan\left(\cos^{-1}\left(\dfrac{3}{5}\right)\right) = \dfrac{4}{3}

Your Turn 1.4

Using exact values only, with no calculator, evaluate $\tan\left(\cos^{-1}\left(- \dfrac{12}{13}\right)\right)$ .

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Colour keybasic angle or value found on the wayfinal value

Let

\theta = \cos^{-1}\left(- \dfrac{12}{13}\right)

\displaystyle \cos\theta\;= - \dfrac{12}{13}

The principal value of

\cos^{-1}

is an angle on the upper half of the unit circle, so

\sin\theta \ge 0

\displaystyle \sin\theta\;= \sqrt{1 - \left(- \dfrac{12}{13}\right)^2} = \sqrt{\dfrac{25}{169}} = \textcolor{#010105}{\dfrac{5}{13}}

Pythagorean identity, positive root because

\sin\theta \ge 0

on this range.

\displaystyle \tan\theta\;= \dfrac{\sin\theta}{\cos\theta} = \dfrac{\dfrac{5}{13}}{- \dfrac{12}{13}} = \textcolor{#010106}{- \dfrac{5}{12}}

Quotient identity.

Answer

\tan\left(\cos^{-1}\left(- \dfrac{12}{13}\right)\right) = - \dfrac{5}{12}

Example 1.5

Cosine of an inverse tangent of a negative number

Using exact values only, with no calculator, evaluate $\cos\left(\tan^{-1}\left(- \dfrac{3}{4}\right)\right)$ .

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Colour keybasic angle or value found on the way

Let

\theta = \tan^{-1}\left(- \dfrac{3}{4}\right)

\displaystyle \tan\theta\;= - \dfrac{3}{4}

The principal value of

\tan^{-1}

is an angle strictly on the right-hand half of the unit circle, so

\cos\theta > 0

\displaystyle \sec^{2}\theta\;= 1 + \tan^{2}\theta = 1 + \left(- \dfrac{3}{4}\right)^{2} = \textcolor{#010105}{\dfrac{25}{16}}

Pythagorean identity

\sec^2\theta = 1 + \tan^2\theta

\displaystyle \cos\theta\;= \sqrt{\dfrac{1}{\dfrac{25}{16}}} = \textcolor{#010105}{\dfrac{4}{5}}

\cos\theta = \dfrac{1}{\sec\theta}

, positive on this range.

\displaystyle \sin\theta\;= \tan\theta \cos\theta = \left(- \dfrac{3}{4}\right)\left(\dfrac{4}{5}\right) = \textcolor{#010105}{- \dfrac{3}{5}}

From

\tan\theta = \dfrac{\sin\theta}{\cos\theta}

Answer

\cos\left(\tan^{-1}\left(- \dfrac{3}{4}\right)\right) = \dfrac{4}{5}

Your Turn 1.5

Using exact values only, with no calculator, evaluate $\sin\left(\tan^{-1}\left(\dfrac{5}{12}\right)\right)$ .

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Colour keybasic angle or value found on the way

Let

\theta = \tan^{-1}\left(\dfrac{5}{12}\right)

\displaystyle \tan\theta\;= \dfrac{5}{12}

The principal value of

\tan^{-1}

is an angle strictly on the right-hand half of the unit circle, so

\cos\theta > 0

\displaystyle \sec^{2}\theta\;= 1 + \tan^{2}\theta = 1 + \left(\dfrac{5}{12}\right)^{2} = \textcolor{#010105}{\dfrac{169}{144}}

Pythagorean identity

\sec^2\theta = 1 + \tan^2\theta

\displaystyle \cos\theta\;= \sqrt{\dfrac{1}{\dfrac{169}{144}}} = \textcolor{#010105}{\dfrac{12}{13}}

\cos\theta = \dfrac{1}{\sec\theta}

, positive on this range.

\displaystyle \sin\theta\;= \tan\theta \cos\theta = \left(\dfrac{5}{12}\right)\left(\dfrac{12}{13}\right) = \textcolor{#010105}{\dfrac{5}{13}}

From

\tan\theta = \dfrac{\sin\theta}{\cos\theta}

Answer

\sin\left(\tan^{-1}\left(\dfrac{5}{12}\right)\right) = \dfrac{5}{13}

Links To

The signs by quadrant, related angles and the graphs of the six ratios are in Chapter 9: Trigonometric functions and graphs

Supports K341-G1-06

Compound angle formulae and exact values

What is $\sin 75^{\circ}$ exactly? It is not on the special-angle table, but $75^{\circ} = 45^{\circ} + 30^{\circ}$ and both of those are. The compound-angle formulae turn the sine, cosine or tangent of a sum or a difference into ratios of the two parts.

The Core Idea

Compound-angle formulae (on the formula page).

$\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B$ , $\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B$ and $\tan(A \pm B) = \dfrac{\tan A \pm \tan B}{1 \mp \tan A\tan B}$ . They hold for every pair of angles, in degrees or in radians.

The six compound-angle formulae and the sign rule
Formula	Expansion	Sign rule	Picture
$\sin(A + B)$	$\sin A\cos B + \cos A\sin B$	same sign as inside	The dashed segment is sin(A+B).
$\sin(A - B)$	$\sin A\cos B - \cos A\sin B$	same sign as inside	The dashed segment is sin(A−B).
$\cos(A + B)$	$\cos A\cos B - \sin A\sin B$	opposite sign to inside	The dashed segment is cos(A+B).
$\cos(A - B)$	$\cos A\cos B + \sin A\sin B$	opposite sign to inside	The dashed segment is cos(A−B).
$\tan(A + B)$	$\dfrac{\tan A + \tan B}{1 - \tan A\tan B}$	denominator has the opposite sign	The dashed segment is tan(A+B).
$\tan(A - B)$	$\dfrac{\tan A - \tan B}{1 + \tan A\tan B}$	denominator has the opposite sign	The dashed segment is tan(A−B).

SGSchoolKaki Tip

Cosine flips the sign: $\cos(A + B)$ has a minus between the products and $\cos(A - B)$ has a plus. Sine keeps the sign. For tangent the sign in the denominator is the opposite of the sign inside the bracket.

Example 2.1

Exact value: sine of a sum

Use a compound-angle formula to find the exact value of $\sin 75^{\circ}$ .

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Colour keyfinal value

\displaystyle \sin\left(45^{\circ} + 30^{\circ}\right)\;= \sin 45^{\circ} \cos 30^{\circ} + \cos 45^{\circ} \sin 30^{\circ}

Sine of a sum:

\sin(A + B) = \sin A \cos B + \cos A \sin B

\displaystyle = \left(\dfrac{\sqrt{2}}{2}\right)\left(\dfrac{\sqrt{3}}{2}\right) + \left(\dfrac{\sqrt{2}}{2}\right)\left(\dfrac{1}{2}\right)

Replace each ratio by its exact value from the special-angle triangles.

\displaystyle = \dfrac{\sqrt{6}}{4} + \dfrac{\sqrt{2}}{4}

Multiply out each product and keep each term in surd form.

\displaystyle = \textcolor{#010106}{\dfrac{\sqrt{2} + \sqrt{6}}{4}}

Combine into one fraction (rationalising if needed).

Answer

\sin\left(45^{\circ} + 30^{\circ}\right) = \dfrac{\sqrt{2} + \sqrt{6}}{4}

Your Turn 2.1

Use a compound-angle formula to find the exact value of $\sin 15^{\circ}$ .

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Colour keyfinal value

\displaystyle \sin\left(60^{\circ} - 45^{\circ}\right)\;= \sin 60^{\circ} \cos 45^{\circ} - \cos 60^{\circ} \sin 45^{\circ}

Sine of a difference:

\sin(A - B) = \sin A \cos B - \cos A \sin B

\displaystyle = \left(\dfrac{\sqrt{3}}{2}\right)\left(\dfrac{\sqrt{2}}{2}\right) - \left(\dfrac{1}{2}\right)\left(\dfrac{\sqrt{2}}{2}\right)

Replace each ratio by its exact value from the special-angle triangles.

\displaystyle = \dfrac{\sqrt{6}}{4} - \dfrac{\sqrt{2}}{4}

Multiply out each product and keep each term in surd form.

\displaystyle = \textcolor{#010106}{\dfrac{\sqrt{6} - \sqrt{2}}{4}}

Combine into one fraction (rationalising if needed).

Answer

\sin\left(60^{\circ} - 45^{\circ}\right) = \dfrac{\sqrt{6} - \sqrt{2}}{4}

Example 2.2

Exact value: cosine of a sum

Use a compound-angle formula to find the exact value of $\cos 105^{\circ}$ .

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Colour keyfinal value

\displaystyle \cos\left(60^{\circ} + 45^{\circ}\right)\;= \cos 60^{\circ} \cos 45^{\circ} - \sin 60^{\circ} \sin 45^{\circ}

Cosine of a sum: the sign between the two products is the opposite of the sign inside:

\cos(A + B) = \cos A \cos B - \sin A \sin B

\displaystyle = \left(\dfrac{1}{2}\right)\left(\dfrac{\sqrt{2}}{2}\right) - \left(\dfrac{\sqrt{3}}{2}\right)\left(\dfrac{\sqrt{2}}{2}\right)

Replace each ratio by its exact value from the special-angle triangles.

\displaystyle = \dfrac{\sqrt{2}}{4} - \dfrac{\sqrt{6}}{4}

Multiply out each product and keep each term in surd form.

\displaystyle = \textcolor{#010106}{\dfrac{\sqrt{2} - \sqrt{6}}{4}}

Combine into one fraction (rationalising if needed).

Answer

\cos\left(60^{\circ} + 45^{\circ}\right) = \dfrac{\sqrt{2} - \sqrt{6}}{4}

Your Turn 2.2

Use a compound-angle formula to find the exact value of $\cos 75^{\circ}$ .

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Colour keyfinal value

\displaystyle \cos\left(45^{\circ} + 30^{\circ}\right)\;= \cos 45^{\circ} \cos 30^{\circ} - \sin 45^{\circ} \sin 30^{\circ}

Cosine of a sum: the sign between the two products is the opposite of the sign inside:

\cos(A + B) = \cos A \cos B - \sin A \sin B

\displaystyle = \left(\dfrac{\sqrt{2}}{2}\right)\left(\dfrac{\sqrt{3}}{2}\right) - \left(\dfrac{\sqrt{2}}{2}\right)\left(\dfrac{1}{2}\right)

Replace each ratio by its exact value from the special-angle triangles.

\displaystyle = \dfrac{\sqrt{6}}{4} - \dfrac{\sqrt{2}}{4}

Multiply out each product and keep each term in surd form.

\displaystyle = \textcolor{#010106}{\dfrac{\sqrt{6} - \sqrt{2}}{4}}

Combine into one fraction (rationalising if needed).

Answer

\cos\left(45^{\circ} + 30^{\circ}\right) = \dfrac{\sqrt{6} - \sqrt{2}}{4}

\sin(A + B)

is not

\sin A + \sin B

, and

\tan 2A

is not

2\tan A

. Test with

A = B = 30^{\circ}

\sin 60^{\circ} = \dfrac{\sqrt{3}}{2} \approx 0.87

but

\sin 30^{\circ} + \sin 30^{\circ} = 1

. In general a function of a sum is not the sum of the functions: always use the formula.

Example 2.3

Exact value: tangent of a difference or a sum

Use a compound-angle formula to find the exact value of $\tan 15^{\circ}$ .

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Colour keyfinal value

\displaystyle \tan\left(60^{\circ} - 45^{\circ}\right)\;= \dfrac{\tan 60^{\circ} - \tan 45^{\circ}}{1 + \tan 60^{\circ} \tan 45^{\circ}}

Tangent of a difference:

\tan(A - B) = \dfrac{\tan A - \tan B}{1 + \tan A \tan B}

\displaystyle = \dfrac{\sqrt{3} - 1}{1 + \left(\sqrt{3}\right)1}

Replace each ratio by its exact value from the special-angle triangles.

\displaystyle = \dfrac{-1 + \sqrt{3}}{1 + \sqrt{3}}

Multiply out each product and keep each term in surd form.

\displaystyle = \textcolor{#010106}{2 - \sqrt{3}}

Divide, then rationalise the denominator by multiplying top and bottom by its conjugate.

Answer

\tan\left(60^{\circ} - 45^{\circ}\right) = 2 - \sqrt{3}

Your Turn 2.3

Use a compound-angle formula to find the exact value of $\tan 105^{\circ}$ .

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Colour keyfinal value

\displaystyle \tan\left(60^{\circ} + 45^{\circ}\right)\;= \dfrac{\tan 60^{\circ} + \tan 45^{\circ}}{1 - \tan 60^{\circ} \tan 45^{\circ}}

Tangent of a sum:

\tan(A + B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B}

; the sign in the denominator is the opposite.

\displaystyle = \dfrac{\sqrt{3} + 1}{1 - \left(\sqrt{3}\right)1}

Replace each ratio by its exact value from the special-angle triangles.

\displaystyle = \dfrac{1 + \sqrt{3}}{1 - \sqrt{3}}

Multiply out each product and keep each term in surd form.

\displaystyle = \textcolor{#010106}{-2 - \sqrt{3}}

Divide, then rationalise the denominator by multiplying top and bottom by its conjugate.

Answer

\tan\left(60^{\circ} + 45^{\circ}\right) = -2 - \sqrt{3}

Example 2.4

Exact value in radians

Use a compound-angle formula to find the exact value of $\sin \dfrac{\pi}{12}$ .

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Colour keyfinal value

\displaystyle \sin\left(\dfrac{\pi}{4} - \dfrac{\pi}{6}\right)\;= \sin \dfrac{\pi}{4} \cos \dfrac{\pi}{6} - \cos \dfrac{\pi}{4} \sin \dfrac{\pi}{6}

Sine of a difference:

\sin(A - B) = \sin A \cos B - \cos A \sin B

\displaystyle = \left(\dfrac{\sqrt{2}}{2}\right)\left(\dfrac{\sqrt{3}}{2}\right) - \left(\dfrac{\sqrt{2}}{2}\right)\left(\dfrac{1}{2}\right)

Replace each ratio by its exact value from the special-angle triangles.

\displaystyle = \dfrac{\sqrt{6}}{4} - \dfrac{\sqrt{2}}{4}

Multiply out each product and keep each term in surd form.

\displaystyle = \textcolor{#010106}{\dfrac{\sqrt{6} - \sqrt{2}}{4}}

Combine into one fraction (rationalising if needed).

Answer

\sin\left(\dfrac{\pi}{4} - \dfrac{\pi}{6}\right) = \dfrac{\sqrt{6} - \sqrt{2}}{4}

Your Turn 2.4

Use a compound-angle formula to find the exact value of $\cos \dfrac{7\pi}{12}$ .

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Colour keyfinal value

\displaystyle \cos\left(\dfrac{\pi}{3} + \dfrac{\pi}{4}\right)\;= \cos \dfrac{\pi}{3} \cos \dfrac{\pi}{4} - \sin \dfrac{\pi}{3} \sin \dfrac{\pi}{4}

Cosine of a sum: the sign between the two products is the opposite of the sign inside:

\cos(A + B) = \cos A \cos B - \sin A \sin B

\displaystyle = \left(\dfrac{1}{2}\right)\left(\dfrac{\sqrt{2}}{2}\right) - \left(\dfrac{\sqrt{3}}{2}\right)\left(\dfrac{\sqrt{2}}{2}\right)

Replace each ratio by its exact value from the special-angle triangles.

\displaystyle = \dfrac{\sqrt{2}}{4} - \dfrac{\sqrt{6}}{4}

Multiply out each product and keep each term in surd form.

\displaystyle = \textcolor{#010106}{\dfrac{\sqrt{2} - \sqrt{6}}{4}}

Combine into one fraction (rationalising if needed).

Answer

\cos\left(\dfrac{\pi}{3} + \dfrac{\pi}{4}\right) = \dfrac{\sqrt{2} - \sqrt{6}}{4}

Think About It

\cos(A + B)

equal to

\cos A + \cos B

? Test

A = B = 60^{\circ}

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\displaystyle \cos\left(60^{\circ} + 60^{\circ}\right)\;= \cos 120^{\circ} = -\dfrac{1}{2}

Add the angles first, then take the cosine:

120^{\circ}

is in the second quadrant, where cosine is negative.

\displaystyle \cos 60^{\circ} + \cos 60^{\circ}\;= \dfrac{1}{2} + \dfrac{1}{2} = 1

Take each cosine first, then add: a positive number.

Answer

-\dfrac{1}{2} \ne 1

, so

\cos(A + B)

is not

\cos A + \cos B

. The correct expansion is

\cos A\cos B - \sin A\sin B = \dfrac{1}{4} - \dfrac{3}{4} = -\dfrac{1}{2}

When $A$ and $B$ are not special angles but you are given a ratio of each and the quadrant of each, first find the missing sine and cosine of each angle with the Pythagorean identity (the quadrant gives the signs), then use the formula.

Example 2.5

Two angles, sum of sine and cosine

The angle $A$ lies in the first quadrant and the angle $B$ lies in the first quadrant. Given that $\sin A = \dfrac{4}{5}$ and $\cos B = \dfrac{5}{13}$ , find the exact values of $\sin(A + B)$ and $\cos(A + B)$ .

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Colour keybasic angle or value found on the wayfinal value

\displaystyle \cos^{2} A\;= 1 - \sin^{2} A = 1 - \left(\dfrac{4}{5}\right)^{2} = \textcolor{#010105}{\dfrac{9}{25}}

Pythagorean identity:

\sin^2 A + \cos^2 A = 1

, rearranged.

\displaystyle \cos A\;= \sqrt{\dfrac{9}{25}} = \textcolor{#010105}{\dfrac{3}{5}}

Angle A is in quadrant I, where cosine is positive, so take the positive root.

\displaystyle \tan A\;= \dfrac{\sin A}{\cos A} = \dfrac{\dfrac{4}{5}}{\dfrac{3}{5}} = \textcolor{#010105}{\dfrac{4}{3}}

Quotient identity:

\tan A = \dfrac{\sin A}{\cos A}

\displaystyle \sin^{2} B\;= 1 - \cos^{2} B = 1 - \left(\dfrac{5}{13}\right)^{2} = \textcolor{#010105}{\dfrac{144}{169}}

Pythagorean identity:

\sin^2 A + \cos^2 A = 1

, rearranged.

\displaystyle \sin B\;= \sqrt{\dfrac{144}{169}} = \textcolor{#010105}{\dfrac{12}{13}}

Angle B is in quadrant I, where sine is positive, so take the positive root.

\displaystyle \tan B\;= \dfrac{\sin B}{\cos B} = \dfrac{\dfrac{12}{13}}{\dfrac{5}{13}} = \textcolor{#010105}{\dfrac{12}{5}}

Quotient identity:

\tan A = \dfrac{\sin A}{\cos A}

\sin(A + B) = \sin A \cos B + \cos A \sin B

Compound-angle formula for sine.

\displaystyle = \left(\dfrac{4}{5}\right)\left(\dfrac{5}{13}\right) + \left(\dfrac{3}{5}\right)\left(\dfrac{12}{13}\right) = \textcolor{#010106}{\dfrac{56}{65}}

Substitute the four values found above.

\cos(A + B) = \cos A \cos B - \sin A \sin B

Compound-angle formula for cosine: the sign between the products is opposite to the sign inside the bracket.

\displaystyle = \left(\dfrac{3}{5}\right)\left(\dfrac{5}{13}\right) - \left(\dfrac{4}{5}\right)\left(\dfrac{12}{13}\right) = \textcolor{#010106}{- \dfrac{33}{65}}

Substitute the four values found above.

Answer

\sin(A + B) = \dfrac{56}{65}

;

\cos(A + B) = - \dfrac{33}{65}

Your Turn 2.5

The angle $A$ lies in the second quadrant and the angle $B$ lies in the first quadrant. Given that $\sin A = \dfrac{3}{5}$ and $\cos B = \dfrac{8}{17}$ , find the exact values of $\sin(A + B)$ and $\cos(A + B)$ .

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Colour keybasic angle or value found on the wayfinal value

\displaystyle \cos^{2} A\;= 1 - \sin^{2} A = 1 - \left(\dfrac{3}{5}\right)^{2} = \textcolor{#010105}{\dfrac{16}{25}}

Pythagorean identity:

\sin^2 A + \cos^2 A = 1

, rearranged.

\displaystyle \cos A\;= -\sqrt{\dfrac{16}{25}} = \textcolor{#010105}{- \dfrac{4}{5}}

Angle A is in quadrant II, where cosine is negative, so take the negative root.

\displaystyle \tan A\;= \dfrac{\sin A}{\cos A} = \dfrac{\dfrac{3}{5}}{- \dfrac{4}{5}} = \textcolor{#010105}{- \dfrac{3}{4}}

Quotient identity:

\tan A = \dfrac{\sin A}{\cos A}

\displaystyle \sin^{2} B\;= 1 - \cos^{2} B = 1 - \left(\dfrac{8}{17}\right)^{2} = \textcolor{#010105}{\dfrac{225}{289}}

Pythagorean identity:

\sin^2 A + \cos^2 A = 1

, rearranged.

\displaystyle \sin B\;= \sqrt{\dfrac{225}{289}} = \textcolor{#010105}{\dfrac{15}{17}}

Angle B is in quadrant I, where sine is positive, so take the positive root.

\displaystyle \tan B\;= \dfrac{\sin B}{\cos B} = \dfrac{\dfrac{15}{17}}{\dfrac{8}{17}} = \textcolor{#010105}{\dfrac{15}{8}}

Quotient identity:

\tan A = \dfrac{\sin A}{\cos A}

\sin(A + B) = \sin A \cos B + \cos A \sin B

Compound-angle formula for sine.

\displaystyle = \left(\dfrac{3}{5}\right)\left(\dfrac{8}{17}\right) + \left(- \dfrac{4}{5}\right)\left(\dfrac{15}{17}\right) = \textcolor{#010106}{- \dfrac{36}{85}}

Substitute the four values found above.

\cos(A + B) = \cos A \cos B - \sin A \sin B

Compound-angle formula for cosine: the sign between the products is opposite to the sign inside the bracket.

\displaystyle = \left(- \dfrac{4}{5}\right)\left(\dfrac{8}{17}\right) - \left(\dfrac{3}{5}\right)\left(\dfrac{15}{17}\right) = \textcolor{#010106}{- \dfrac{77}{85}}

Substitute the four values found above.

Answer

\sin(A + B) = - \dfrac{36}{85}

;

\cos(A + B) = - \dfrac{77}{85}

Example 2.6

Tangent of a difference, angles in quadrants IV and II

The angle $A$ lies in the fourth quadrant and the angle $B$ lies in the second quadrant. Given that $\tan A = - \dfrac{5}{12}$ and $\sin B = \dfrac{3}{5}$ , find the exact value of $\tan(A - B)$ .

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Colour keybasic angle or value found on the wayfinal value

\displaystyle \sec^{2} A\;= 1 + \tan^{2} A = 1 + \left(- \dfrac{5}{12}\right)^{2} = \textcolor{#010105}{\dfrac{169}{144}}

Pythagorean identity:

\sec^2 A = 1 + \tan^2 A

\displaystyle \cos A\;= \sqrt{\dfrac{1}{\dfrac{169}{144}}} = \textcolor{#010105}{\dfrac{12}{13}}

Angle A is in quadrant IV, where cosine is positive;

\cos^2 A = \dfrac{1}{\sec^2 A}

\displaystyle \sin A\;= \tan A \cos A = \left(- \dfrac{5}{12}\right)\left(\dfrac{12}{13}\right) = \textcolor{#010105}{- \dfrac{5}{13}}

From

\tan A = \dfrac{\sin A}{\cos A}

, so

\sin A = \tan A \cos A

\displaystyle \cos^{2} B\;= 1 - \sin^{2} B = 1 - \left(\dfrac{3}{5}\right)^{2} = \textcolor{#010105}{\dfrac{16}{25}}

Pythagorean identity:

\sin^2 A + \cos^2 A = 1

, rearranged.

\displaystyle \cos B\;= -\sqrt{\dfrac{16}{25}} = \textcolor{#010105}{- \dfrac{4}{5}}

Angle B is in quadrant II, where cosine is negative, so take the negative root.

\displaystyle \tan B\;= \dfrac{\sin B}{\cos B} = \dfrac{\dfrac{3}{5}}{- \dfrac{4}{5}} = \textcolor{#010105}{- \dfrac{3}{4}}

Quotient identity:

\tan A = \dfrac{\sin A}{\cos A}

\tan(A - B) = \dfrac{\tan A - \tan B}{1 + \tan A \tan B}

Compound-angle formula for tangent.

\displaystyle = \dfrac{\left(- \dfrac{5}{12}\right) - \left(- \dfrac{3}{4}\right)}{1 + \left(- \dfrac{5}{12}\right)\left(- \dfrac{3}{4}\right)} = \textcolor{#010106}{\dfrac{16}{63}}

Substitute the two tangents found above.

Answer

\tan(A - B) = \dfrac{16}{63}

Your Turn 2.6

The angle $A$ lies in the first quadrant and the angle $B$ lies in the third quadrant. Given that $\tan A = \dfrac{7}{24}$ and $\sin B = - \dfrac{4}{5}$ , find the exact value of $\tan(A - B)$ .

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Colour keybasic angle or value found on the wayfinal value

\displaystyle \sec^{2} A\;= 1 + \tan^{2} A = 1 + \left(\dfrac{7}{24}\right)^{2} = \textcolor{#010105}{\dfrac{625}{576}}

Pythagorean identity:

\sec^2 A = 1 + \tan^2 A

\displaystyle \cos A\;= \sqrt{\dfrac{1}{\dfrac{625}{576}}} = \textcolor{#010105}{\dfrac{24}{25}}

Angle A is in quadrant I, where cosine is positive;

\cos^2 A = \dfrac{1}{\sec^2 A}

\displaystyle \sin A\;= \tan A \cos A = \left(\dfrac{7}{24}\right)\left(\dfrac{24}{25}\right) = \textcolor{#010105}{\dfrac{7}{25}}

From

\tan A = \dfrac{\sin A}{\cos A}

, so

\sin A = \tan A \cos A

\displaystyle \cos^{2} B\;= 1 - \sin^{2} B = 1 - \left(- \dfrac{4}{5}\right)^{2} = \textcolor{#010105}{\dfrac{9}{25}}

Pythagorean identity:

\sin^2 A + \cos^2 A = 1

, rearranged.

\displaystyle \cos B\;= -\sqrt{\dfrac{9}{25}} = \textcolor{#010105}{- \dfrac{3}{5}}

Angle B is in quadrant III, where cosine is negative, so take the negative root.

\displaystyle \tan B\;= \dfrac{\sin B}{\cos B} = \dfrac{- \dfrac{4}{5}}{- \dfrac{3}{5}} = \textcolor{#010105}{\dfrac{4}{3}}

Quotient identity:

\tan A = \dfrac{\sin A}{\cos A}

\tan(A - B) = \dfrac{\tan A - \tan B}{1 + \tan A \tan B}

Compound-angle formula for tangent.

\displaystyle = \dfrac{\left(\dfrac{7}{24}\right) - \left(\dfrac{4}{3}\right)}{1 + \left(\dfrac{7}{24}\right)\left(\dfrac{4}{3}\right)} = \textcolor{#010106}{- \dfrac{3}{4}}

Substitute the two tangents found above.

Answer

\tan(A - B) = - \dfrac{3}{4}

Example 2.7

Both ratios are cosines

The angle $A$ lies in the second quadrant and the angle $B$ lies in the fourth quadrant. Given that $\cos A = - \dfrac{15}{17}$ and $\cos B = \dfrac{3}{5}$ , find the exact values of $\cos(A - B)$ and $\sin(A - B)$ .

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Colour keybasic angle or value found on the wayfinal value

\displaystyle \sin^{2} A\;= 1 - \cos^{2} A = 1 - \left(- \dfrac{15}{17}\right)^{2} = \textcolor{#010105}{\dfrac{64}{289}}

Pythagorean identity:

\sin^2 A + \cos^2 A = 1

, rearranged.

\displaystyle \sin A\;= \sqrt{\dfrac{64}{289}} = \textcolor{#010105}{\dfrac{8}{17}}

Angle A is in quadrant II, where sine is positive, so take the positive root.

\displaystyle \tan A\;= \dfrac{\sin A}{\cos A} = \dfrac{\dfrac{8}{17}}{- \dfrac{15}{17}} = \textcolor{#010105}{- \dfrac{8}{15}}

Quotient identity:

\tan A = \dfrac{\sin A}{\cos A}

\displaystyle \sin^{2} B\;= 1 - \cos^{2} B = 1 - \left(\dfrac{3}{5}\right)^{2} = \textcolor{#010105}{\dfrac{16}{25}}

Pythagorean identity:

\sin^2 A + \cos^2 A = 1

, rearranged.

\displaystyle \sin B\;= -\sqrt{\dfrac{16}{25}} = \textcolor{#010105}{- \dfrac{4}{5}}

Angle B is in quadrant IV, where sine is negative, so take the negative root.

\displaystyle \tan B\;= \dfrac{\sin B}{\cos B} = \dfrac{- \dfrac{4}{5}}{\dfrac{3}{5}} = \textcolor{#010105}{- \dfrac{4}{3}}

Quotient identity:

\tan A = \dfrac{\sin A}{\cos A}

\cos(A - B) = \cos A \cos B + \sin A \sin B

Compound-angle formula for cosine: the sign between the products is opposite to the sign inside the bracket.

\displaystyle = \left(- \dfrac{15}{17}\right)\left(\dfrac{3}{5}\right) + \left(\dfrac{8}{17}\right)\left(- \dfrac{4}{5}\right) = \textcolor{#010106}{- \dfrac{77}{85}}

Substitute the four values found above.

\sin(A - B) = \sin A \cos B - \cos A \sin B

Compound-angle formula for sine.

\displaystyle = \left(\dfrac{8}{17}\right)\left(\dfrac{3}{5}\right) - \left(- \dfrac{15}{17}\right)\left(- \dfrac{4}{5}\right) = \textcolor{#010106}{- \dfrac{36}{85}}

Substitute the four values found above.

Answer

\cos(A - B) = - \dfrac{77}{85}

;

\sin(A - B) = - \dfrac{36}{85}

Your Turn 2.7

The angle $A$ lies in the third quadrant and the angle $B$ lies in the first quadrant. Given that $\cos A = - \dfrac{7}{25}$ and $\cos B = \dfrac{4}{5}$ , find the exact values of $\cos(A - B)$ and $\sin(A - B)$ .

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Colour keybasic angle or value found on the wayfinal value

\displaystyle \sin^{2} A\;= 1 - \cos^{2} A = 1 - \left(- \dfrac{7}{25}\right)^{2} = \textcolor{#010105}{\dfrac{576}{625}}

Pythagorean identity:

\sin^2 A + \cos^2 A = 1

, rearranged.

\displaystyle \sin A\;= -\sqrt{\dfrac{576}{625}} = \textcolor{#010105}{- \dfrac{24}{25}}

Angle A is in quadrant III, where sine is negative, so take the negative root.

\displaystyle \tan A\;= \dfrac{\sin A}{\cos A} = \dfrac{- \dfrac{24}{25}}{- \dfrac{7}{25}} = \textcolor{#010105}{\dfrac{24}{7}}

Quotient identity:

\tan A = \dfrac{\sin A}{\cos A}

\displaystyle \sin^{2} B\;= 1 - \cos^{2} B = 1 - \left(\dfrac{4}{5}\right)^{2} = \textcolor{#010105}{\dfrac{9}{25}}

Pythagorean identity:

\sin^2 A + \cos^2 A = 1

, rearranged.

\displaystyle \sin B\;= \sqrt{\dfrac{9}{25}} = \textcolor{#010105}{\dfrac{3}{5}}

Angle B is in quadrant I, where sine is positive, so take the positive root.

\displaystyle \tan B\;= \dfrac{\sin B}{\cos B} = \dfrac{\dfrac{3}{5}}{\dfrac{4}{5}} = \textcolor{#010105}{\dfrac{3}{4}}

Quotient identity:

\tan A = \dfrac{\sin A}{\cos A}

\cos(A - B) = \cos A \cos B + \sin A \sin B

Compound-angle formula for cosine: the sign between the products is opposite to the sign inside the bracket.

\displaystyle = \left(- \dfrac{7}{25}\right)\left(\dfrac{4}{5}\right) + \left(- \dfrac{24}{25}\right)\left(\dfrac{3}{5}\right) = \textcolor{#010106}{- \dfrac{4}{5}}

Substitute the four values found above.

\sin(A - B) = \sin A \cos B - \cos A \sin B

Compound-angle formula for sine.

\displaystyle = \left(- \dfrac{24}{25}\right)\left(\dfrac{4}{5}\right) - \left(- \dfrac{7}{25}\right)\left(\dfrac{3}{5}\right) = \textcolor{#010106}{- \dfrac{3}{5}}

Substitute the four values found above.

Answer

\cos(A - B) = - \dfrac{4}{5}

;

\sin(A - B) = - \dfrac{3}{5}

Exam questions often give the ratios as letters instead of numbers: $\sin A = p$ and $\cos B = q$ . The method is identical, and the answer is an expression in $p$ and $q$ . The quadrant still decides each sign, now in front of a square root such as $\sqrt{1 - p^{2}}$ .

Example 2.8

Letters p and q

The angle $A$ lies in the first quadrant and the angle $B$ lies in the first quadrant. Given that $\sin A = p$ and $\cos B = q$ , find $\sin(A + B)$ in terms of $p$ and $q$ .

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Colour keyfinal value

\displaystyle p\;= \sin A

The letter

p

stands for the given ratio

\sin A

\displaystyle \cos^{2} A\;= 1 - \sin^{2} A = 1 - p^{2}

Pythagorean identity:

\sin^2 A + \cos^2 A = 1

, rearranged.

\displaystyle \cos A\;= \sqrt{1 - p^2}

Angle

A

is in quadrant I, where cosine is positive, so take the positive root.

\displaystyle q\;= \cos B

The letter

q

stands for the given ratio

\cos B

\displaystyle \sin^{2} B\;= 1 - \cos^{2} B = 1 - q^{2}

Pythagorean identity:

\sin^2 A + \cos^2 A = 1

, rearranged.

\displaystyle \sin B\;= \sqrt{1 - q^2}

Angle

B

is in quadrant I, where sine is positive, so take the positive root.

\sin(A + B) = \sin A\cos B + \cos A\sin B

Compound-angle formula for sine.

\displaystyle = p \times q + \sqrt{1 - p^2} \times \sqrt{1 - q^2} = \textcolor{#010106}{p q + \sqrt{1 - p^2} \sqrt{1 - q^2}}

Substitute the four expressions found above.

Answer

\sin(A + B) = p q + \sqrt{1 - p^2} \sqrt{1 - q^2}

Your Turn 2.8

The angle $A$ lies in the first quadrant and the angle $B$ lies in the first quadrant. Given that $\cos A = p$ and $\tan B = q$ , find $\cos(A - B)$ in terms of $p$ and $q$ .

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Colour keyfinal value

\displaystyle p\;= \cos A

The letter

p

stands for the given ratio

\cos A

\displaystyle \sin^{2} A\;= 1 - \cos^{2} A = 1 - p^{2}

Pythagorean identity:

\sin^2 A + \cos^2 A = 1

, rearranged.

\displaystyle \sin A\;= \sqrt{1 - p^2}

Angle

A

is in quadrant I, where sine is positive, so take the positive root.

\displaystyle q\;= \tan B

The letter

q

stands for the given ratio

\tan B

\displaystyle \sec^{2} B\;= 1 + \tan^{2} B = 1 + q^{2}

Pythagorean identity:

\sec^2 A = 1 + \tan^2 A

\displaystyle \cos B\;= \dfrac{1}{\sqrt{1 + q^2}}

Angle

B

is in quadrant I, where cosine is positive;

\cos A = \dfrac{1}{\sec A}

with the sign of the quadrant.

\sin B = \tan B\cos B

From

\tan A = \dfrac{\sin A}{\cos A}

, so

\sin A = \tan A \cos A

\displaystyle \sin B\;= \dfrac{q}{\sqrt{q^2 + 1}}

Substitute

\tan A

and the cosine found above.

\cos(A - B) = \cos A\cos B + \sin A\sin B

Compound-angle formula for cosine: the sign between the products is opposite to the sign inside the bracket.

\displaystyle = p \times \dfrac{1}{\sqrt{q^2 + 1}} + \sqrt{1 - p^2} \times \dfrac{q}{\sqrt{q^2 + 1}} = \textcolor{#010106}{\dfrac{p}{\sqrt{q^2 + 1}} + \dfrac{q \sqrt{1 - p^2}}{\sqrt{q^2 + 1}}}

Substitute the four expressions found above.

Answer

\cos(A - B) = \dfrac{p}{\sqrt{q^2 + 1}} + \dfrac{q \sqrt{1 - p^2}}{\sqrt{q^2 + 1}}

Links To

Related angles and the signs in each quadrant, used throughout this section, are in Chapter 9: Trigonometric functions and graphs

Supports K341-G1-06 · K341-G1-07

Double angle and half angle

Put $B = A$ in the compound-angle formulae and the double-angle formulae appear. They are on the formula page, and the half-angle results used in this chapter come straight from them.

The Core Idea

Double-angle formulae (on the formula page).

$\sin 2A = 2\sin A\cos A$ , $\cos 2A = \cos^{2}A - \sin^{2}A = 2\cos^{2}A - 1 = 1 - 2\sin^{2}A$ and $\tan 2A = \dfrac{2\tan A}{1 - \tan^{2}A}$ . Rearranging the last two forms of $\cos 2A$ gives the half-angle results $\cos^{2}A = \dfrac{1 + \cos 2A}{2}$ and $\sin^{2}A = \dfrac{1 - \cos 2A}{2}$ , which are not on the page: memorise them.

Which form of cos 2A to use
You are given	Use	Why	Picture
$\sin A$ only	$1 - 2\sin^{2}A$	only sine appears	Only the height (sine) of A is known.
$\cos A$ only	$2\cos^{2}A - 1$	only cosine appears	Only the width (cosine) of A is known.
$\sin A$ and $\cos A$ (or $\tan A$ with the quadrant)	$\cos^{2}A - \sin^{2}A$	both are known	Both the height and the width of A are known.

SGSchoolKaki Tip

Choose the form of $\cos 2A$ that uses the ratio you were given. Given $\sin A$ , use $1 - 2\sin^{2}A$ : no square root and no quadrant sign is needed for $\cos 2A$ .

Example 3.1

All three double-angle ratios from the sine

Given that $\sin A = \dfrac{5}{13}$ and that $A$ lies in the second quadrant, find the exact values of $\sin 2A$ , $\cos 2A$ and $\tan 2A$ .

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Colour keybasic angle or value found on the wayfinal value

\displaystyle \cos^{2} A\;= 1 - \sin^{2} A = 1 - \left(\dfrac{5}{13}\right)^{2} = \textcolor{#010105}{\dfrac{144}{169}}

Pythagorean identity:

\sin^2 A + \cos^2 A = 1

, rearranged.

\displaystyle \cos A\;= -\sqrt{\dfrac{144}{169}} = \textcolor{#010105}{- \dfrac{12}{13}}

Angle A is in quadrant II, where cosine is negative, so take the negative root.

\displaystyle \tan A\;= \dfrac{\sin A}{\cos A} = \dfrac{\dfrac{5}{13}}{- \dfrac{12}{13}} = \textcolor{#010105}{- \dfrac{5}{12}}

Quotient identity:

\tan A = \dfrac{\sin A}{\cos A}

\sin 2A = 2 \sin A \cos A

Double-angle formula for sine.

\displaystyle = 2\left(\dfrac{5}{13}\right)\left(- \dfrac{12}{13}\right) = \textcolor{#010106}{- \dfrac{120}{169}}

Substitute the two ratios found above.

\cos 2A = 1 - 2\sin^{2} A

Of the three forms of

\cos 2A

, choose the one that uses the ratio you were given: here

\sin

\displaystyle = 1 - 2\left(\dfrac{5}{13}\right)^{2} = \textcolor{#010106}{\dfrac{119}{169}}

Substitute the sine found above.

\tan 2A = \dfrac{2\tan A}{1 - \tan^{2} A}

Double-angle formula for tangent.

\displaystyle = \dfrac{2\left(- \dfrac{5}{12}\right)}{1 - \left(- \dfrac{5}{12}\right)^{2}} = \textcolor{#010106}{- \dfrac{120}{119}}

Substitute the tangent found above.

Answer

\sin 2A = - \dfrac{120}{169}

;

\cos 2A = \dfrac{119}{169}

;

\tan 2A = - \dfrac{120}{119}

Your Turn 3.1

Given that $\sin A = - \dfrac{3}{5}$ and that $A$ lies in the fourth quadrant, find the exact values of $\sin 2A$ , $\cos 2A$ and $\tan 2A$ .

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Colour keybasic angle or value found on the wayfinal value

\displaystyle \cos^{2} A\;= 1 - \sin^{2} A = 1 - \left(- \dfrac{3}{5}\right)^{2} = \textcolor{#010105}{\dfrac{16}{25}}

Pythagorean identity:

\sin^2 A + \cos^2 A = 1

, rearranged.

\displaystyle \cos A\;= \sqrt{\dfrac{16}{25}} = \textcolor{#010105}{\dfrac{4}{5}}

Angle A is in quadrant IV, where cosine is positive, so take the positive root.

\displaystyle \tan A\;= \dfrac{\sin A}{\cos A} = \dfrac{- \dfrac{3}{5}}{\dfrac{4}{5}} = \textcolor{#010105}{- \dfrac{3}{4}}

Quotient identity:

\tan A = \dfrac{\sin A}{\cos A}

\sin 2A = 2 \sin A \cos A

Double-angle formula for sine.

\displaystyle = 2\left(- \dfrac{3}{5}\right)\left(\dfrac{4}{5}\right) = \textcolor{#010106}{- \dfrac{24}{25}}

Substitute the two ratios found above.

\cos 2A = 1 - 2\sin^{2} A

Of the three forms of

\cos 2A

, choose the one that uses the ratio you were given: here

\sin

\displaystyle = 1 - 2\left(- \dfrac{3}{5}\right)^{2} = \textcolor{#010106}{\dfrac{7}{25}}

Substitute the sine found above.

\tan 2A = \dfrac{2\tan A}{1 - \tan^{2} A}

Double-angle formula for tangent.

\displaystyle = \dfrac{2\left(- \dfrac{3}{4}\right)}{1 - \left(- \dfrac{3}{4}\right)^{2}} = \textcolor{#010106}{- \dfrac{24}{7}}

Substitute the tangent found above.

Answer

\sin 2A = - \dfrac{24}{25}

;

\cos 2A = \dfrac{7}{25}

;

\tan 2A = - \dfrac{24}{7}

Example 3.2

Cosine and tangent of 2A from the tangent

Given that $\tan A = - \dfrac{4}{3}$ and that $A$ lies in the second quadrant, find the exact values of $\cos 2A$ and $\tan 2A$ .

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Colour keybasic angle or value found on the wayfinal value

\displaystyle \sec^{2} A\;= 1 + \tan^{2} A = 1 + \left(- \dfrac{4}{3}\right)^{2} = \textcolor{#010105}{\dfrac{25}{9}}

Pythagorean identity:

\sec^2 A = 1 + \tan^2 A

\displaystyle \cos A\;= -\sqrt{\dfrac{1}{\dfrac{25}{9}}} = \textcolor{#010105}{- \dfrac{3}{5}}

Angle A is in quadrant II, where cosine is negative;

\cos^2 A = \dfrac{1}{\sec^2 A}

\displaystyle \sin A\;= \tan A \cos A = \left(- \dfrac{4}{3}\right)\left(- \dfrac{3}{5}\right) = \textcolor{#010105}{\dfrac{4}{5}}

From

\tan A = \dfrac{\sin A}{\cos A}

, so

\sin A = \tan A \cos A

\cos 2A = \cos^{2} A - \sin^{2} A

Both

\sin

and

\cos

are known, so use the form

\cos^2 A - \sin^2 A

\displaystyle = \left(- \dfrac{3}{5}\right)^{2} - \left(\dfrac{4}{5}\right)^{2} = \textcolor{#010106}{- \dfrac{7}{25}}

Substitute both ratios found above.

\tan 2A = \dfrac{2\tan A}{1 - \tan^{2} A}

Double-angle formula for tangent.

\displaystyle = \dfrac{2\left(- \dfrac{4}{3}\right)}{1 - \left(- \dfrac{4}{3}\right)^{2}} = \textcolor{#010106}{\dfrac{24}{7}}

Substitute the tangent found above.

Answer

\cos 2A = - \dfrac{7}{25}

;

\tan 2A = \dfrac{24}{7}

Your Turn 3.2

Given that $\tan A = \dfrac{5}{12}$ and that $A$ lies in the first quadrant, find the exact values of $\cos 2A$ and $\tan 2A$ .

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Colour keybasic angle or value found on the wayfinal value

\displaystyle \sec^{2} A\;= 1 + \tan^{2} A = 1 + \left(\dfrac{5}{12}\right)^{2} = \textcolor{#010105}{\dfrac{169}{144}}

Pythagorean identity:

\sec^2 A = 1 + \tan^2 A

\displaystyle \cos A\;= \sqrt{\dfrac{1}{\dfrac{169}{144}}} = \textcolor{#010105}{\dfrac{12}{13}}

Angle A is in quadrant I, where cosine is positive;

\cos^2 A = \dfrac{1}{\sec^2 A}

\displaystyle \sin A\;= \tan A \cos A = \left(\dfrac{5}{12}\right)\left(\dfrac{12}{13}\right) = \textcolor{#010105}{\dfrac{5}{13}}

From

\tan A = \dfrac{\sin A}{\cos A}

, so

\sin A = \tan A \cos A

\cos 2A = \cos^{2} A - \sin^{2} A

Both

\sin

and

\cos

are known, so use the form

\cos^2 A - \sin^2 A

\displaystyle = \left(\dfrac{12}{13}\right)^{2} - \left(\dfrac{5}{13}\right)^{2} = \textcolor{#010106}{\dfrac{119}{169}}

Substitute both ratios found above.

\tan 2A = \dfrac{2\tan A}{1 - \tan^{2} A}

Double-angle formula for tangent.

\displaystyle = \dfrac{2\left(\dfrac{5}{12}\right)}{1 - \left(\dfrac{5}{12}\right)^{2}} = \textcolor{#010106}{\dfrac{120}{119}}

Substitute the tangent found above.

Answer

\cos 2A = \dfrac{119}{169}

;

\tan 2A = \dfrac{120}{119}

\tan 2A

is not

2\tan A

, and

\sin 2A

is not

2\sin A

. Test with

A = 45^{\circ}

\sin 90^{\circ} = 1

but

2\sin 45^{\circ} = \sqrt{2}

Example 3.3

Half angle: the angle x is in the first two quadrants

Given that $\cos x = \dfrac{7}{25}$ and $0^{\circ} < x < 180^{\circ}$ , find the exact values of $\cos^{2}\dfrac{x}{2}$ and $\cos\dfrac{x}{2}$ .

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Colour keyfinal value

\displaystyle \cos^{2}\dfrac{x}{2}\;= \dfrac{1 + \cos x}{2} = \dfrac{1 + \left(\dfrac{7}{25}\right)}{2} = \textcolor{#010106}{\dfrac{16}{25}}

From

\cos 2A = 2\cos^2 A - 1

with

A = \dfrac{x}{2}

\cos^2\dfrac{x}{2} = \dfrac{1 + \cos x}{2}

Since

0^{\circ} < x < 180^{\circ}

, the half angle satisfies

0^{\circ} < \dfrac{x}{2} < 90^{\circ}

, which is in quadrant I, where cos is positive.

\displaystyle \cos\dfrac{x}{2}\;= \sqrt{\dfrac{16}{25}} = \textcolor{#010106}{\dfrac{4}{5}}

Take the positive square root because of the quadrant.

Answer

\cos^{2}\dfrac{x}{2} = \dfrac{16}{25}

;

\cos\dfrac{x}{2} = \dfrac{4}{5}

Your Turn 3.3

Given that $\cos x = - \dfrac{7}{25}$ and $0^{\circ} < x < 180^{\circ}$ , find the exact values of $\cos^{2}\dfrac{x}{2}$ and $\cos\dfrac{x}{2}$ .

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Colour keyfinal value

\displaystyle \cos^{2}\dfrac{x}{2}\;= \dfrac{1 + \cos x}{2} = \dfrac{1 + \left(- \dfrac{7}{25}\right)}{2} = \textcolor{#010106}{\dfrac{9}{25}}

From

\cos 2A = 2\cos^2 A - 1

with

A = \dfrac{x}{2}

\cos^2\dfrac{x}{2} = \dfrac{1 + \cos x}{2}

Since

0^{\circ} < x < 180^{\circ}

, the half angle satisfies

0^{\circ} < \dfrac{x}{2} < 90^{\circ}

, which is in quadrant I, where cos is positive.

\displaystyle \cos\dfrac{x}{2}\;= \sqrt{\dfrac{9}{25}} = \textcolor{#010106}{\dfrac{3}{5}}

Take the positive square root because of the quadrant.

Answer

\cos^{2}\dfrac{x}{2} = \dfrac{9}{25}

;

\cos\dfrac{x}{2} = \dfrac{3}{5}

Example 3.4

Half angle: a surd answer

Given that $\cos x = - \dfrac{1}{9}$ and $0^{\circ} < x < 180^{\circ}$ , find the exact values of $\sin^{2}\dfrac{x}{2}$ and $\sin\dfrac{x}{2}$ .

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Colour keyfinal value

\displaystyle \sin^{2}\dfrac{x}{2}\;= \dfrac{1 - \cos x}{2} = \dfrac{1 - \left(- \dfrac{1}{9}\right)}{2} = \textcolor{#010106}{\dfrac{5}{9}}

From

\cos 2A = 1 - 2\sin^2 A

with

A = \dfrac{x}{2}

\sin^2\dfrac{x}{2} = \dfrac{1 - \cos x}{2}

Since

0^{\circ} < x < 180^{\circ}

, the half angle satisfies

0^{\circ} < \dfrac{x}{2} < 90^{\circ}

, which is in quadrant I, where sin is positive.

\displaystyle \sin\dfrac{x}{2}\;= \sqrt{\dfrac{5}{9}} = \textcolor{#010106}{\dfrac{\sqrt{5}}{3}}

Take the positive square root because of the quadrant.

Answer

\sin^{2}\dfrac{x}{2} = \dfrac{5}{9}

;

\sin\dfrac{x}{2} = \dfrac{\sqrt{5}}{3}

Your Turn 3.4

Given that $\cos x = \dfrac{17}{81}$ and $0^{\circ} < x < 180^{\circ}$ , find the exact values of $\sin^{2}\dfrac{x}{2}$ and $\sin\dfrac{x}{2}$ .

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Colour keyfinal value

\displaystyle \sin^{2}\dfrac{x}{2}\;= \dfrac{1 - \cos x}{2} = \dfrac{1 - \left(\dfrac{17}{81}\right)}{2} = \textcolor{#010106}{\dfrac{32}{81}}

From

\cos 2A = 1 - 2\sin^2 A

with

A = \dfrac{x}{2}

\sin^2\dfrac{x}{2} = \dfrac{1 - \cos x}{2}

Since

0^{\circ} < x < 180^{\circ}

, the half angle satisfies

0^{\circ} < \dfrac{x}{2} < 90^{\circ}

, which is in quadrant I, where sin is positive.

\displaystyle \sin\dfrac{x}{2}\;= \sqrt{\dfrac{32}{81}} = \textcolor{#010106}{\dfrac{4 \sqrt{2}}{9}}

Take the positive square root because of the quadrant.

Answer

\sin^{2}\dfrac{x}{2} = \dfrac{32}{81}

;

\sin\dfrac{x}{2} = \dfrac{4 \sqrt{2}}{9}

Think About It

The three forms of

\cos 2A

look different. Show that the second and third forms both come from the first.

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\displaystyle \cos 2A\;= \cos^{2}A - \sin^{2}A

Put

B = A

\cos(A + B) = \cos A\cos B - \sin A\sin B

\displaystyle = \cos^{2}A - \left(1 - \cos^{2}A\right) = 2\cos^{2}A - 1

Replace

\sin^2 A

1 - \cos^2 A

to keep only cosines.

\displaystyle \cos 2A\;= \left(1 - \sin^{2}A\right) - \sin^{2}A = 1 - 2\sin^{2}A

Replace

\cos^2 A

1 - \sin^2 A

to keep only sines.

Answer

One identity,

\sin^{2}A + \cos^{2}A = 1

, turns the first form into the other two. So the three forms are always equal, and you may pick the most convenient.

Example 3.5

Find m: a sum of two sines

Find the exact value of the constant $m$ for which $\sin(x + 30^{\circ}) + \sin(x - 30^{\circ}) = m\sin x$ for every value of $x$ .

Show solutionHide solution

Colour keyfinal value

\sin(x + 30^{\circ}) + \sin(x - 30^{\circ}) = \left(\sin x\cos 30^{\circ} + \cos x\sin 30^{\circ}\right) + \left(\sin x\cos 30^{\circ} - \cos x\sin 30^{\circ}\right)

Expand both brackets with the compound-angle formula for sine.

\displaystyle = 2\sin x\cos 30^{\circ}

The

\cos x \sin a

terms are equal and opposite, so they cancel; the

\sin x \cos a

terms add.

\displaystyle = 2\left(\dfrac{\sqrt{3}}{2}\right)\sin x = \textcolor{#010106}{\sqrt{3}\sin x}

Replace the exact value of the constant.

Answer

m = \sqrt{3}

Your Turn 3.5

Find the exact value of the constant $m$ for which $\cos(x + 45^{\circ}) + \cos(x - 45^{\circ}) = m\cos x$ for every value of $x$ .

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Colour keyfinal value

\cos(x + 45^{\circ}) + \cos(x - 45^{\circ}) = \left(\cos x\cos 45^{\circ} - \sin x\sin 45^{\circ}\right) + \left(\cos x\cos 45^{\circ} + \sin x\sin 45^{\circ}\right)

Expand both brackets with the compound-angle formula for cosine.

\displaystyle = 2\cos x\cos 45^{\circ}

The

\sin x \sin a

terms are equal and opposite, so they cancel; the

\cos x \cos a

terms add.

\displaystyle = 2\left(\dfrac{\sqrt{2}}{2}\right)\cos x = \textcolor{#010106}{\sqrt{2}\cos x}

Replace the exact value of the constant.

Answer

m = \sqrt{2}

Example 3.6

Cosine of twice an inverse sine

Using exact values only, with no calculator, evaluate $\cos\left(2\sin^{-1}\left(\dfrac{3}{5}\right)\right)$ .

Show solutionHide solution

Colour keybasic angle or value found on the wayfinal value

Let

\theta = \sin^{-1}\left(\dfrac{3}{5}\right)

\displaystyle \sin\theta\;= \dfrac{3}{5}

The principal value

\theta = \sin^{-1}\left(\dfrac{3}{5}\right)

is an angle on the right-hand half of the unit circle, so

\cos\theta \ge 0

\displaystyle \cos\theta\;= \sqrt{1 - \left(\dfrac{3}{5}\right)^2} = \sqrt{\dfrac{16}{25}} = \textcolor{#010105}{\dfrac{4}{5}}

Pythagorean identity, positive root because

\cos\theta \ge 0

on this range.

\displaystyle \cos 2\theta\;= \cos^{2}\theta - \sin^{2}\theta = \left(\dfrac{4}{5}\right)^{2} - \left(\dfrac{3}{5}\right)^{2} = \textcolor{#010106}{\dfrac{7}{25}}

Double-angle formula for cosine, in the form that uses both ratios.

Answer

\cos\left(2\sin^{-1}\left(\dfrac{3}{5}\right)\right) = \dfrac{7}{25}

Your Turn 3.6

Using exact values only, with no calculator, evaluate $\tan\left(2\cos^{-1}\left(\dfrac{4}{5}\right)\right)$ .

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Colour keybasic angle or value found on the wayfinal value

Let

\theta = \cos^{-1}\left(\dfrac{4}{5}\right)

\displaystyle \cos\theta\;= \dfrac{4}{5}

The principal value of

\cos^{-1}

is an angle on the upper half of the unit circle, so

\sin\theta \ge 0

\displaystyle \sin\theta\;= \sqrt{1 - \left(\dfrac{4}{5}\right)^2} = \sqrt{\dfrac{9}{25}} = \textcolor{#010105}{\dfrac{3}{5}}

Pythagorean identity, positive root because

\sin\theta \ge 0

on this range.

\displaystyle \tan 2\theta\;= \dfrac{2\tan\theta}{1 - \tan^{2}\theta} = \dfrac{2\left(\dfrac{3}{4}\right)}{1 - \left(\dfrac{3}{4}\right)^{2}} = \textcolor{#010106}{\dfrac{24}{7}}

Double-angle formula for tangent, with

\tan\theta = \dfrac{\sin\theta}{\cos\theta}

found above.

Answer

\tan\left(2\cos^{-1}\left(\dfrac{4}{5}\right)\right) = \dfrac{24}{7}

Supports K341-G1-07 · K341-G1-09

Simplifying and proving identities

An identity is true for every angle where both sides exist, so it is proved, not solved. The aim is to start from one side and change it, one true step at a time, until it becomes the other side. The skill is choosing which step to take, and a short list of habits covers almost every proof you will meet.

The Core Idea

Prove an identity by working on one side only.

Start with the left side and turn it into the right side. Never start from the result: writing the answer first and then changing both sides assumes what you are trying to show. Four moves do most of the work: write everything in sine and cosine, put fractions over a common denominator, multiply by the conjugate (the same bracket with the middle sign changed, so that $(1 - \cos x)(1 + \cos x) = 1 - \cos^{2} x = \sin^{2} x$ ), and use a double-angle formula that removes the 1 or the square. A proof has at most six short lines.

Which technique for which look?
Technique	Use it when	The move	Picture
Sine and cosine	the identity has $\tan$ , $\cot$ , $\sec$ or $\operatorname{cosec}$	$\tan x = \dfrac{\sin x}{\cos x}$ , $\sec x = \dfrac{1}{\cos x}$ , $\operatorname{cosec} x = \dfrac{1}{\sin x}$	Both sides trace one curve: tan x cos x = sin x.
Common denominator	two fractions are added or subtracted	combine them, then use $\sin^2 x + \cos^2 x = 1$ on the top or the bottom	Both sides trace one curve: 1/(2 + sin x) + 1/(2 − sin x) = 4/(4 − sin²x).
Conjugate	a denominator or numerator looks like $1 \pm \sin x$ or $1 \pm \cos x$	multiply top and bottom by the same bracket with the sign changed	Both sides trace one curve: (1 − cos x)(1 + cos x) = sin²x.
Double angle	$2x$ appears, or a $1 \pm \cos 2x$	pick the form of $\cos 2x$ that cancels the 1: $1 + \cos 2x = 2\cos^2 x$ , $1 - \cos 2x = 2\sin^2 x$	Both sides trace one curve: 1 + cos 2x = 2cos²x.
Square then expand	a bracket is squared, or a difference of two squares appears	expand, then use $\sin^2 x + \cos^2 x = 1$ and $\sin 2x = 2\sin x\cos x$	Both sides trace one curve: (sin x + cos x)² = 1 + sin 2x.

Example 4.1

Turn everything into sine and cosine

Prove that $\tan x + \cot x \equiv \sec x\operatorname{cosec} x$ .

Show solutionHide solution

Colour keyfinal value

\tan x + \cot x = \dfrac{\sin x}{\cos x} + \dfrac{\cos x}{\sin x}

Rewrite

\tan

and

\cot

as sine over cosine.

\displaystyle = \dfrac{\sin^{2} x + \cos^{2} x}{\sin x\cos x}

Put both fractions over the common denominator

\sin\cos

\displaystyle = \dfrac{1}{\sin x\cos x}

Pythagorean identity:

\sin^2 + \cos^2 = 1

\displaystyle = \dfrac{1}{\cos x} \times \dfrac{1}{\sin x} = \textcolor{#010106}{\sec x\operatorname{cosec} x}

Split into two reciprocals.

Answer

\tan x + \cot x \equiv \sec x\operatorname{cosec} x

(proved)

Your Turn 4.1

Prove that $\operatorname{cosec} x - \sin x \equiv \cos x\cot x$ .

Show full solutionHide solution

Colour keyfinal value

\operatorname{cosec} x - \sin x = \dfrac{1}{\sin x} - \sin x

Write

\operatorname{cosec}

as 1 over

\sin

\displaystyle = \dfrac{1 - \sin^{2} x}{\sin x}

Common denominator

\sin

\displaystyle = \dfrac{\cos^{2} x}{\sin x}

Pythagorean identity:

1 - \sin^2 = \cos^2

\displaystyle = \cos x \times \dfrac{\cos x}{\sin x} = \textcolor{#010106}{\cos x\cot x}

Split off one

\cos

and use

\cot = \cos / \sin

Answer

\operatorname{cosec} x - \sin x \equiv \cos x\cot x

(proved)

SGSchoolKaki Tip

Turn everything into sine and cosine first. Tangent, cotangent, secant and cosecant are all fractions of sine and cosine, so after one step you are working with just two functions and the Pythagorean identity does the rest.

Example 4.2

Add the fractions over a common denominator

Prove that $\dfrac{1}{1 - \sin x} + \dfrac{1}{1 + \sin x} \equiv 2\sec^{2} x$ .

Show solutionHide solution

Colour keyfinal value

\dfrac{1}{1 - \sin x} + \dfrac{1}{1 + \sin x} = \dfrac{\left(1 + \sin x\right) + \left(1 - \sin x\right)}{\left(1 - \sin x\right)\left(1 + \sin x\right)}

Common denominator: multiply each fraction's top and bottom by the other bracket.

\displaystyle = \dfrac{2}{1 - \sin^{2} x}

The

\sin

terms cancel on top; the bottom is a difference of two squares.

\displaystyle = \dfrac{2}{\cos^{2} x}

Pythagorean identity:

1 - \sin^2 = \cos^2

\displaystyle = \textcolor{#010106}{2\sec^{2} x}

\dfrac{1}{\cos^2} = \sec^2

Answer

\dfrac{1}{1 - \sin x} + \dfrac{1}{1 + \sin x} \equiv 2\sec^{2} x

(proved)

Your Turn 4.2

Prove that $\left(\operatorname{cosec} x + \cot x\right)\left(1 - \cos x\right) \equiv \sin x$ .

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Colour keyfinal value

\left(\operatorname{cosec} x + \cot x\right)\left(1 - \cos x\right) = \left(\dfrac{1}{\sin x} + \dfrac{\cos x}{\sin x}\right)\left(1 - \cos x\right)

Rewrite

\operatorname{cosec}

and

\cot

in sine and cosine.

\displaystyle = \dfrac{\left(1 + \cos x\right)\left(1 - \cos x\right)}{\sin x}

Add the fractions, then multiply by the bracket.

\displaystyle = \dfrac{1 - \cos^{2} x}{\sin x}

Difference of two squares.

\displaystyle = \dfrac{\sin^{2} x}{\sin x} = \textcolor{#010106}{\sin x}

Pythagorean identity, then cancel one

\sin

Answer

\left(\operatorname{cosec} x + \cot x\right)\left(1 - \cos x\right) \equiv \sin x

(proved)

Example 4.3

Multiply by the conjugate

Prove that $\dfrac{1 - \cos x}{\sin x} \equiv \dfrac{\sin x}{1 + \cos x}$ .

Show solutionHide solution

Colour keyfinal value

\dfrac{1 - \cos x}{\sin x} = \dfrac{\left(1 - \cos x\right)\left(1 + \cos x\right)}{\sin x\left(1 + \cos x\right)}

Multiply top and bottom by the conjugate

1 + \cos

to create a difference of squares.

\displaystyle = \dfrac{1 - \cos^{2} x}{\sin x\left(1 + \cos x\right)}

Difference of two squares on top.

\displaystyle = \dfrac{\sin^{2} x}{\sin x\left(1 + \cos x\right)}

Pythagorean identity:

1 - \cos^2 = \sin^2

\displaystyle = \textcolor{#010106}{\dfrac{\sin x}{1 + \cos x}}

Cancel one factor of

\sin

from top and bottom.

Answer

\dfrac{1 - \cos x}{\sin x} \equiv \dfrac{\sin x}{1 + \cos x}

(proved)

Your Turn 4.3

Prove that $\dfrac{\sin x}{1 - \cos x} \equiv \dfrac{1 + \cos x}{\sin x}$ .

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Colour keyfinal value

\dfrac{\sin x}{1 - \cos x} = \dfrac{\sin x\left(1 + \cos x\right)}{\left(1 - \cos x\right)\left(1 + \cos x\right)}

Multiply top and bottom by the conjugate

1 + \cos

\displaystyle = \dfrac{\sin x\left(1 + \cos x\right)}{1 - \cos^{2} x}

Difference of two squares in the denominator.

\displaystyle = \dfrac{\sin x\left(1 + \cos x\right)}{\sin^{2} x}

Pythagorean identity:

1 - \cos^2 = \sin^2

\displaystyle = \textcolor{#010106}{\dfrac{1 + \cos x}{\sin x}}

Cancel one factor of

\sin

Answer

\dfrac{\sin x}{1 - \cos x} \equiv \dfrac{1 + \cos x}{\sin x}

(proved)

Do not move terms across the identity sign. Adding something to both sides or cross-multiplying is a way of solving an equation, and it assumes the two sides are already equal. In a proof the two sides are only shown equal at the very end, so you may only change one side, using steps that are always true.

Example 4.4

Use the double-angle form that removes the 1

Prove that $\dfrac{\sin 2x}{1 + \cos 2x} \equiv \tan x$ .

Show solutionHide solution

Colour keyfinal value

\dfrac{\sin 2x}{1 + \cos 2x} = \dfrac{2\sin x\cos x}{1 + \left(2\cos^{2} x - 1\right)}

Double-angle formulae:

\sin 2A = 2\sin A\cos A

and the form

\cos 2A = 2\cos^2 A - 1

(it removes the 1 in the denominator).

\displaystyle = \dfrac{2\sin x\cos x}{2\cos^{2} x}

The

1

and

-1

cancel.

\displaystyle = \dfrac{\sin x}{\cos x} = \textcolor{#010106}{\tan x}

Cancel

2\cos

from top and bottom.

Answer

\dfrac{\sin 2x}{1 + \cos 2x} \equiv \tan x

(proved)

Your Turn 4.4

Prove that $\dfrac{1 - \cos 2x}{\sin 2x} \equiv \tan x$ .

Show full solutionHide solution

Colour keyfinal value

\dfrac{1 - \cos 2x}{\sin 2x} = \dfrac{1 - \left(1 - 2\sin^{2} x\right)}{2\sin x\cos x}

Double-angle formulae: the form

\cos 2A = 1 - 2\sin^2 A

removes the 1, and

\sin 2A = 2\sin A\cos A

\displaystyle = \dfrac{2\sin^{2} x}{2\sin x\cos x}

The

1

and

-1

cancel.

\displaystyle = \dfrac{\sin x}{\cos x} = \textcolor{#010106}{\tan x}

Cancel

2\sin

from top and bottom.

Answer

\dfrac{1 - \cos 2x}{\sin 2x} \equiv \tan x

(proved)

Think About It

A student writes

\sqrt{\cos^{2} x} = \cos x

. Is this always true? Test it with an angle in the second quadrant,

x = 120^{\circ}

Show the full workingHide the working

\displaystyle \cos 120^{\circ}\;= -\cos 60^{\circ} = -\dfrac{1}{2}

Second quadrant: cosine is negative, and the basic angle is

180^{\circ} - 120^{\circ} = 60^{\circ}

\displaystyle \cos^{2} 120^{\circ}\;= \left(-\dfrac{1}{2}\right)^{2} = \dfrac{1}{4}

Squaring removes the minus sign.

\displaystyle \sqrt{\cos^{2} 120^{\circ}}\;= \sqrt{\dfrac{1}{4}} = \dfrac{1}{2}

A square root is never negative.

Answer

The square root gives

\dfrac{1}{2}

but

\cos 120^{\circ} = -\dfrac{1}{2}

, so

\sqrt{\cos^{2} x} = \cos x

is false here. In general

\sqrt{\cos^{2} x} = |\cos x|

, which equals

\cos x

only when

\cos x

is not negative, for example for an acute angle. So the root of a square needs the quadrant of the angle.

Example 4.5

Find the constant m

Given that $\sin(x + 60^{\circ}) - \sin(x - 60^{\circ}) = m\cos x$ , find the exact value of $m$ .

Show solutionHide solution

Colour keyfinal value

\sin(x + 60^{\circ}) - \sin(x - 60^{\circ}) = \left(\sin x\cos 60^{\circ} + \cos x\sin 60^{\circ}\right) - \left(\sin x\cos 60^{\circ} - \cos x\sin 60^{\circ}\right)

Expand both brackets with the compound-angle formula for sine.

\displaystyle = 2\cos x\sin 60^{\circ}

The

\sin x \cos a

terms cancel; the

\cos x \sin a

terms add.

\displaystyle = 2\left(\dfrac{\sqrt{3}}{2}\right)\cos x = \textcolor{#010106}{\sqrt{3}\cos x}

Replace the exact value of the constant.

Answer

m = \sqrt{3}

Your Turn 4.5

Given that $\cos(x - 45^{\circ}) - \cos(x + 45^{\circ}) = m\sin x$ , find the exact value of $m$ .

Show full solutionHide solution

Colour keyfinal value

\cos(x - 45^{\circ}) - \cos(x + 45^{\circ}) = \left(\cos x\cos 45^{\circ} + \sin x\sin 45^{\circ}\right) - \left(\cos x\cos 45^{\circ} - \sin x\sin 45^{\circ}\right)

Expand both brackets with the compound-angle formula for cosine.

\displaystyle = 2\sin x\sin 45^{\circ}

The

\cos x \cos a

terms cancel; the

\sin x \sin a

terms add.

\displaystyle = 2\left(\dfrac{\sqrt{2}}{2}\right)\sin x = \textcolor{#010106}{\sqrt{2}\sin x}

Replace the exact value of the constant.

Answer

m = \sqrt{2}

Links To

After a proof comes the other half of the skill: a proved identity is a tool for solving equations. The next section, Solving trigonometric equations in a given interval, shows how to find every angle in a stated range, and the next two Examples already use it with the word hence. Chapter 10: Trigonometric identities, equations and proofs

Example 4.6

Show that, then hence solve

Consider the expression $\cos^{4} x - \sin^{4} x$ .

(i)Prove that $\cos^{4} x - \sin^{4} x \equiv \cos 2x$ .
(ii)Hence solve the equation $\cos^{4} x - \sin^{4} x = \dfrac{1}{2}$ for $0^{\circ} \le x \le 360^{\circ}$ , giving each answer exactly, or correct to 1 decimal place if it is not exact.

Show solutionHide solution

Colour keybasic angle or value found on the wayfinal value

(i)

\cos^{4} x - \sin^{4} x = \left(\cos^{2} x - \sin^{2} x\right)\left(\cos^{2} x + \sin^{2} x\right)

Difference of two squares, with

\cos^4 = (\cos^2)^2

\displaystyle = \left(\cos^{2} x - \sin^{2} x\right) \times 1

Pythagorean identity:

\cos^2 + \sin^2 = 1

\displaystyle = \textcolor{#010106}{\cos 2x}

Double-angle formula

\cos 2A = \cos^2 A - \sin^2 A

(ii)

\cos^{4} x - \sin^{4} x = \cos 2x

Use the identity proved in the previous part: the left side of the equation is the left side of that identity, so it can be replaced by the right side.

\displaystyle \cos 2x\;= \dfrac{1}{2}

Write the equation again with the right side of the identity.

Let

\theta = 2x

. Since

0^{\circ} \le x \le 360^{\circ}

, the angle

\theta

runs from

0^{\circ}

720^{\circ}

, so list every angle with this ratio in that range.

\displaystyle \text{largest angle}\;= 2 \times 360^{\circ} = \textcolor{#010105}{720^{\circ}}

Largest angle: put the largest

x

into

\theta

\displaystyle \alpha\;= \cos^{-1}\left(\dfrac{1}{2}\right) = \textcolor{#010105}{60^{\circ}}

The basic angle

\alpha

is the acute angle with this ratio: use the inverse function on the positive value.

\displaystyle \theta_1\;= \textcolor{#010105}{60^{\circ}}

Since cosine is positive in quadrants I and IV, the angles with this ratio in one full turn are:

\displaystyle \theta_2\;= 360^{\circ} - 60^{\circ} = \textcolor{#010105}{300^{\circ}}

The other angle in the same turn, from the symmetry of the quadrants.

\displaystyle \theta_3\;= 60^{\circ} + 360^{\circ} = \textcolor{#010105}{420^{\circ}}

Add a full turn to

\theta_1

: the ratio repeats every full turn.

\displaystyle \theta_4\;= 300^{\circ} + 360^{\circ} = \textcolor{#010105}{660^{\circ}}

Add a full turn to

\theta_2

: the ratio repeats every full turn.

\displaystyle x\;= \dfrac{60^{\circ}}{2} = \textcolor{#010106}{30^{\circ}}

Divide by the coefficient of

x

to return from

\theta

x

\displaystyle x\;= \dfrac{300^{\circ}}{2} = \textcolor{#010106}{150^{\circ}}

Divide by the coefficient of

x

to return from

\theta

x

\displaystyle x\;= \dfrac{420^{\circ}}{2} = \textcolor{#010106}{210^{\circ}}

Divide by the coefficient of

x

to return from

\theta

x

\displaystyle x\;= \dfrac{660^{\circ}}{2} = \textcolor{#010106}{330^{\circ}}

Divide by the coefficient of

x

to return from

\theta

x

Answer

(i)

\cos^{4} x - \sin^{4} x \equiv \cos 2x

(proved)

(ii)

x = 30^{\circ}, 150^{\circ}, 210^{\circ}, 330^{\circ}

Your Turn 4.6

Consider the expression $\dfrac{1 - \tan^{2} x}{1 + \tan^{2} x}$ .

(i)Prove that $\dfrac{1 - \tan^{2} x}{1 + \tan^{2} x} \equiv \cos 2x$ .
(ii)Hence solve the equation $\dfrac{1 - \tan^{2} x}{1 + \tan^{2} x} = \dfrac{3}{5}$ for $0^{\circ} \le x \le 360^{\circ}$ , giving each answer exactly, or correct to 1 decimal place if it is not exact.

Show full solutionHide solution

Colour keybasic angle or value found on the wayfinal value

(i)

\dfrac{1 - \tan^{2} x}{1 + \tan^{2} x} = \dfrac{1 - \tan^{2} x}{\sec^{2} x}

Pythagorean identity:

1 + \tan^2 = \sec^2

\displaystyle = \left(1 - \dfrac{\sin^{2} x}{\cos^{2} x}\right)\cos^{2} x

Write

\dfrac{1}{\sec^2}

\cos^2

and

\tan

\sin / \cos

\displaystyle = \cos^{2} x - \sin^{2} x = \textcolor{#010106}{\cos 2x}

Multiply out; this is the double-angle formula for cosine.

(ii)

\dfrac{1 - \tan^{2} x}{1 + \tan^{2} x} = \cos 2x

Use the identity proved in the previous part: the left side of the equation is the left side of that identity, so it can be replaced by the right side.

\displaystyle \cos 2x\;= \dfrac{3}{5}

Write the equation again with the right side of the identity.

Let

\theta = 2x

. Since

0^{\circ} \le x \le 360^{\circ}

, the angle

\theta

runs from

0^{\circ}

720^{\circ}

, so list every angle with this ratio in that range.

\displaystyle \text{largest angle}\;= 2 \times 360^{\circ} = \textcolor{#010105}{720^{\circ}}

Largest angle: put the largest

x

into

\theta

\displaystyle \alpha\;= \cos^{-1}\left(\dfrac{3}{5}\right) \approx \textcolor{#010105}{53.13^{\circ}}

The basic angle

\alpha

is the acute angle with this ratio: use the inverse function on the positive value.

\displaystyle \theta_1\;= \textcolor{#010105}{53.13^{\circ}}

Since cosine is positive in quadrants I and IV, the angles with this ratio in one full turn are:

\displaystyle \theta_2\;= 360^{\circ} - 53.13^{\circ} = \textcolor{#010105}{306.87^{\circ}}

The other angle in the same turn, from the symmetry of the quadrants.

\displaystyle \theta_3\;= 53.13^{\circ} + 360^{\circ} = \textcolor{#010105}{413.13^{\circ}}

Add a full turn to

\theta_1

: the ratio repeats every full turn.

\displaystyle \theta_4\;= 306.87^{\circ} + 360^{\circ} = \textcolor{#010105}{666.87^{\circ}}

Add a full turn to

\theta_2

: the ratio repeats every full turn.

\displaystyle x\;= \dfrac{53.13^{\circ}}{2} \approx \textcolor{#010106}{26.6^{\circ}}

Divide by the coefficient of

x

to return from

\theta

x

\displaystyle x\;= \dfrac{306.87^{\circ}}{2} \approx \textcolor{#010106}{153.4^{\circ}}

Divide by the coefficient of

x

to return from

\theta

x

\displaystyle x\;= \dfrac{413.13^{\circ}}{2} \approx \textcolor{#010106}{206.6^{\circ}}

Divide by the coefficient of

x

to return from

\theta

x

\displaystyle x\;= \dfrac{666.87^{\circ}}{2} \approx \textcolor{#010106}{333.4^{\circ}}

Divide by the coefficient of

x

to return from

\theta

x

Answer

(i)

\dfrac{1 - \tan^{2} x}{1 + \tan^{2} x} \equiv \cos 2x

(proved)

(ii)

x = 26.6^{\circ}, 153.4^{\circ}, 206.6^{\circ}, 333.4^{\circ}

Example 4.7

A second show-that-then-hence

Consider the expression $\cot x - \tan x$ .

(i)Prove that $\cot x - \tan x \equiv 2\cot 2x$ .
(ii)Hence solve the equation $\cot x - \tan x = 2$ for $0^{\circ} \le x \le 180^{\circ}$ .

Show solutionHide solution

Colour keybasic angle or value found on the wayfinal value

(i)

\cot x - \tan x = \dfrac{\cos x}{\sin x} - \dfrac{\sin x}{\cos x}

Rewrite

\cot

and

\tan

in sine and cosine.

\displaystyle = \dfrac{\cos^{2} x - \sin^{2} x}{\sin x\cos x}

Common denominator

\sin\cos

\displaystyle = \dfrac{\cos 2x}{\frac{1}{2}\sin 2x}

Double-angle formulae:

\cos^2 - \sin^2 = \cos 2A

and

\sin A\cos A = \frac{1}{2}\sin 2A

\displaystyle = \dfrac{2\cos 2x}{\sin 2x} = \textcolor{#010106}{2\cot 2x}

Clear the half in the denominator.

(ii)

\displaystyle \cot x - \tan x\;= 2\cot 2x

Use the identity proved in the previous part: the left side of the equation is the left side of that identity, so it can be replaced by the right side.

\displaystyle 2\cot 2x\;= 2

Write the equation again with the right side of the identity.

2\cot 2x = 2 \times \dfrac{1}{\tan 2x} = \dfrac{2}{\tan 2x}

\cot

is the reciprocal of

\tan

\displaystyle \dfrac{2}{\tan 2x}\;= 2

The equation in terms of

\tan 2x

\displaystyle \tan 2x\;= \dfrac{2}{2} = 1

Swap

\tan 2x

and the right side: if

\dfrac{2}{\tan 2x}

equals a number, then

\tan 2x

equals

2

divided by that number.

Let

\theta = 2x

. Since

0^{\circ} \le x \le 180^{\circ}

, the angle

\theta

runs from

0^{\circ}

360^{\circ}

, so list every angle with this ratio in that range.

\displaystyle \text{largest angle}\;= 2 \times 180^{\circ} = \textcolor{#010105}{360^{\circ}}

Largest angle: put the largest

x

into

\theta

\displaystyle \alpha\;= \tan^{-1}\left(1\right) = \textcolor{#010105}{45^{\circ}}

The basic angle

\alpha

is the acute angle with this ratio: use the inverse function on the positive value.

\displaystyle \theta_1\;= \textcolor{#010105}{45^{\circ}}

Since tangent is positive in quadrants I and III, and repeats every half turn, the angles with this ratio in one half turn are:

\displaystyle \theta_2\;= 45^{\circ} + 180^{\circ} = \textcolor{#010105}{225^{\circ}}

Add a half turn to

\theta_1

: the ratio repeats every half turn.

\displaystyle x\;= \dfrac{45^{\circ}}{2} = \textcolor{#010106}{22.5^{\circ}}

Divide by the coefficient of

x

to return from

\theta

x

\displaystyle x\;= \dfrac{225^{\circ}}{2} = \textcolor{#010106}{112.5^{\circ}}

Divide by the coefficient of

x

to return from

\theta

x

Answer

(i)

\cot x - \tan x \equiv 2\cot 2x

(proved)

(ii)

x = 22.5^{\circ}, 112.5^{\circ}

Your Turn 4.7

Consider the expression $\dfrac{1 + \cos 2x}{\sin 2x}$ .

(i)Prove that $\dfrac{1 + \cos 2x}{\sin 2x} \equiv \cot x$ .
(ii)Hence solve the equation $\dfrac{1 + \cos 2x}{\sin 2x} = \sqrt{3}$ for $0^{\circ} \le x \le 180^{\circ}$ .

Show full solutionHide solution

Colour keybasic angle or value found on the wayfinal value

(i)

\dfrac{1 + \cos 2x}{\sin 2x} = \dfrac{1 + \left(2\cos^{2} x - 1\right)}{2\sin x\cos x}

Double-angle formulae: the form

\cos 2A = 2\cos^2 A - 1

removes the 1, and

\sin 2A = 2\sin A\cos A

\displaystyle = \dfrac{2\cos^{2} x}{2\sin x\cos x}

The

1

and

-1

cancel.

\displaystyle = \dfrac{\cos x}{\sin x} = \textcolor{#010106}{\cot x}

Cancel

2\cos

from top and bottom.

(ii)

\dfrac{1 + \cos 2x}{\sin 2x} = \cot x

Use the identity proved in the previous part: the left side of the equation is the left side of that identity, so it can be replaced by the right side.

\displaystyle \cot x\;= \sqrt{3}

Write the equation again with the right side of the identity.

\cot x = \dfrac{1}{\tan x}

\cot

is the reciprocal of

\tan

\displaystyle \dfrac{1}{\tan x}\;= \sqrt{3}

The equation in terms of

\tan x

\displaystyle \tan x\;= \dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}

Swap

\tan x

and the right side: if

\dfrac{1}{\tan x}

equals a number, then

\tan x

equals

1

divided by that number.

\displaystyle \alpha\;= \tan^{-1}\left(\dfrac{\sqrt{3}}{3}\right) = \textcolor{#010105}{30^{\circ}}

The basic angle

\alpha

is the acute angle with this ratio: use the inverse function on the positive value.

\displaystyle x_1\;= \textcolor{#010105}{30^{\circ}}

Since tangent is positive in quadrants I and III, and repeats every half turn, the angles with this ratio in one half turn are:

Answer

(i)

\dfrac{1 + \cos 2x}{\sin 2x} \equiv \cot x

(proved)

(ii)

x = 30^{\circ}

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Notes version 1.0 · Last updated 3 Nov 2026 · Chapter 10 of 17