Chapter 9 · Additional Maths · Trigonometric functions and graphs

Trigonometric functions and graphs: angles, radians and waves

Sine, cosine and tangent describe anything that turns or repeats: a Ferris wheel, a searchlight, a tide. This chapter measures angles in degrees and in radians, finds exact values at any angle with the quadrant rules, reads principal values off the unit circle, and then draws and decodes the graphs of y = a sin(bx) + c, y = a cos(bx) + c and y = a tan(bx). Radians start here and every calculus chapter after this one uses them.

Your outcome tracker0 / 7

Let's go! Tick an outcome once you can pass its exam check. Saved on this device.

Chapter
9

Figure 1 · a sine wave in radians: amplitude 2, shifted up 1

5 outcomes · 1 page · every step shown ✦

Radians: π rad = 180°

Multiply degrees by $\dfrac{\pi}{180}$ to get radians. Arc length $s = r\theta$ and sector area $\frac{1}{2}r^{2}\theta$ only work when $\theta$ is in radians.

Quadrant, basic angle, sign

To find a value at any angle, reduce to the basic angle (the acute angle to the $x$ -axis), then use the quadrant to choose the sign: all, sin, tan, cos (ASTC).

a, b, c read the graph

In $y = a\sin(bx) + c$ : amplitude $|a|$ , period $\dfrac{360^{\circ}}{b}$ (or $\dfrac{2\pi}{b}$ ), midline $y = c$ . Tangent has period $\dfrac{180^{\circ}}{b}$ and asymptotes.

By the end of this chapter

Convert between degrees and radians, and use s = rθ and A = ½r²θ for arcs and sectors.
Write the six trigonometric ratios from a right-angled triangle and recall the exact values at 30°, 45° and 60° (π/6, π/4, π/3).
Find the exact value of a trigonometric function at an angle of any size, in degrees or radians, using the quadrant sign and the basic angle.
State the principal value of sin⁻¹, cos⁻¹ and tan⁻¹ for a special value, using the unit circle.
State the amplitude, period, midline, maximum and minimum of y = a sin(bx) + c and y = a cos(bx) + c, and the period and asymptotes of y = a tan(bx).
Sketch these graphs over a given interval, and find a, b and c from a described graph.

✶

Start here

Maths In Real Life

Three situations where an angle drives a height, a distance or an area. The pictures come first; the graph in each solution comes from the same numbers.

$Front view of a Ferris wheel at a school carnival field. A circular wheel with small closed cabins turns on an axle held by an A-frame on level grass, with one cabin at the lowest point just above the ground. A vertical dashed line shows the height of the axle above the ground and a horizontal line shows the radius of the wheel$

14\text{ m}

12\text{ m}

Maths In Real Life 1

A Ferris wheel at the school carnival

A car on a Ferris wheel rises and falls in a smooth repeating wave as the wheel turns, which is exactly the shape of a cosine curve shifted up by the height of the axle.

At a school carnival a Ferris wheel turns steadily. The height $y$ metres of a car above the ground is given by $y = -12\cos x + 14$ , where $x$ degrees is the angle the wheel has turned since the car was at its lowest point.

(a)Find the amplitude and the period of this graph, and the greatest and the least height of the car above the ground.
(b)When $x = 40^{\circ}$ the car is at a certain height. Find every angle $x$ from $0^{\circ}$ to $360^{\circ}$ at which the car is at this same height.

Show full solutionHide solution

Colour keybasic angle, side or quarter periodfinal value

(a)

\displaystyle \text{amplitude}\;= |-12| = \textcolor{#010106}{12}

The amplitude is the size of

a

, ignoring its sign.

\displaystyle \text{period}\;= \textcolor{#010106}{360^{\circ}}

With

b = 1

the curve repeats after one full cycle.

\displaystyle \text{max}\;= 14 + 12 = \textcolor{#010106}{26}

Highest value: midline plus amplitude.

\displaystyle \text{min}\;= 14 - 12 = \textcolor{#010106}{2}

Lowest value: midline minus amplitude.

(b)

\displaystyle \theta_1\;= \textcolor{#010105}{40^{\circ}}

The angle inside the trigonometric function at the given point.

\displaystyle \theta_2\;= 360^{\circ} - 40^{\circ} = \textcolor{#010105}{320^{\circ}}

The cosine curve is symmetric about

0

and

360^{\circ}

\cos\theta = \cos\left(360^{\circ} - \theta\right)

, so the height is the same at

\theta_2

\displaystyle x\;= \textcolor{#010106}{40^{\circ}} \text{ or } \textcolor{#010106}{320^{\circ}}

With

b = 1

the angle is

x

, so both angles inside the turn are solutions.

Answer

(a) amplitude

= 12

; period

= 360^{\circ}

; maximum value

= 26

; minimum value

= 2

(b)

x = 40^{\circ}

320^{\circ}

$A plan view of a small harbour at dusk. A short pier juts up from the bottom edge with a lamp at its tip, a long straight sea wall runs across the top, and calm water lies between them.$

40\text{ m}

d

\theta

\text{plan view}

Maths In Real Life 2

A searchlight sweeping a sea wall

A lighthouse or harbour searchlight turns steadily, but the spot of light it throws on a straight wall does not move steadily: it races along the wall when the beam is nearly parallel to it. That is exactly the behaviour of the tangent curve near an asymptote.

A searchlight stands on a pier, $40$ m from a long straight sea wall, measured along the shortest line to the wall. The beam makes an angle $\theta$ with this shortest line, and the spot of light on the wall is a distance $d$ metres from the nearest point of the wall, where $d = 40\tan\theta$ .

(a)The spot is $\sqrt{3}$ times as far along the wall as the searchlight is from the wall. Find the beam angle $\theta$ , as a principal value.
(b)Write $d$ as $y$ and $\theta$ as $x$ , so that the curve is $y = 40\tan x$ . State its period, and the equations of the two asymptotes that bound the beam angles $-90^{\circ} < x < 90^{\circ}$ .

Show full solutionHide solution

Colour keybasic angle, side or quarter periodfinal value

(a)

The principal value of

\tan^{-1} x

lies in the range

-90^{\circ} < \theta < 90^{\circ}

\displaystyle \tan \theta\;= \sqrt{3}

\tan^{-1}\left(\sqrt{3}\right)

means the angle

\theta

in the principal range whose tan is

\sqrt{3}

\displaystyle \tan 60^{\circ}\;= \sqrt{3}

Reading the special-angle values backwards gives the basic angle.

\displaystyle \theta\;= \textcolor{#010106}{60^{\circ}}

The value is not negative, so the basic angle is already inside the principal range.

(b)

\displaystyle \text{period}\;= \textcolor{#010106}{180^{\circ}}

With

b = 1

the tangent curve repeats every half turn.

The tangent curve has a vertical asymptote wherever the angle

bx

equals an odd multiple of a quarter turn.

\displaystyle \text{first asymptote}\;= \textcolor{#010105}{90^{\circ}}

With

b = 1

the first asymptote on the positive side is one quarter turn from the origin.

\displaystyle \text{gap}\;= 180^{\circ}

Asymptotes repeat every period, so add the period to move to the next one.

\displaystyle x\;= 90^{\circ} - 180^{\circ} = \textcolor{#010105}{-90^{\circ}}

Subtract the gap from the first asymptote to reach the one on the negative side.

\displaystyle x\;= \textcolor{#010106}{-90^{\circ}} \text{ or } \textcolor{#010106}{90^{\circ}}

Answer

(a)

\theta = 60^{\circ}

(b) period

= 180^{\circ}

; asymptotes:

x = -90^{\circ}

90^{\circ}

$A plan view of a perfectly round green lawn in a Singapore estate garden with one small sprinkler head at its centre.$

8\text{ m}

\dfrac{5\pi}{6}

s

\text{plan view}

Maths In Real Life 3

A rotating sprinkler on a round lawn

A rotating sprinkler sweeps back and forth through a fixed angle, so the watered patch is a sector. Because the angle is measured in radians, the length of the curved edge is simply the radius times the angle.

A rotating sprinkler stands at the centre of a round lawn of radius $8$ m. It waters a sector of the lawn by swinging through an angle of $\dfrac{5\pi}{6}$ radians. A fence is to be put along the curved edge of the watered sector.

(a)Express the angle of the sector in degrees.
(b)Find the length of the fence along the curved edge and the area of the lawn that is watered, leaving your answers in terms of $\pi$ .

Show full solutionHide solution

Colour keyfinal value

(a)

\displaystyle \dfrac{5\pi}{6}\text{ rad}\;= \dfrac{5\pi}{6} \times \dfrac{180^{\circ}}{\pi}

Multiply by

\dfrac{180^{\circ}}{\pi}

, the conversion factor from radians to degrees.

\displaystyle = \textcolor{#010106}{150^{\circ}}

The

\pi

cancels, leaving a number of degrees.

(b)

\displaystyle s\;= r\theta = 8 \times \dfrac{5\pi}{6} = \textcolor{#010106}{\dfrac{20\pi}{3}}

Arc length is radius times the angle, and the angle must be in radians.

\displaystyle A\;= \dfrac{1}{2} r^2 \theta = \dfrac{1}{2} \times 8^2 \times \dfrac{5\pi}{6} = \textcolor{#010106}{\dfrac{80\pi}{3}}

Sector area is half the radius squared times the angle in radians.

Answer

(a)

\dfrac{5\pi}{6}

rad

= 150^{\circ}

(b) arc length in m

= \dfrac{20\pi}{3}

; area in square m

= \dfrac{80\pi}{3}

Supports K341-G1-01

Degrees, radians and circles

Until now every angle in this course has been measured in degrees. A second unit, the radian, is used in all the later work on calculus, so this chapter introduces it and then uses both units side by side. One radian is the angle at the centre of a circle when the arc it cuts off is exactly as long as the radius.

Revision. A full turn is

360^{\circ}

, a half turn (angles on a straight line) is

180^{\circ}

and a right angle is

90^{\circ}

. Every conversion in this section rests on these three facts.

Links To

Every calculus chapter that differentiates or integrates a trigonometric function assumes the angle is in radians, starting with Chapter 13: Differentiation of trigonometric, exponential and logarithmic functions

The Core Idea

$\pi$ radians $= 180^{\circ}$ .

A full turn is $2\pi$ radians, so $\pi$ rad $= 180^{\circ}$ . To change degrees into radians, multiply by $\dfrac{\pi}{180}$ . To change radians into degrees, multiply by $\dfrac{180}{\pi}$ . A multiple of $\pi$ such as $\dfrac{5\pi}{6}$ is an exact number of radians; a decimal such as $1.2$ radians is a number of radians too, even with no $\pi$ in sight.

Example 1.1

Degrees to radians

Express $150^{\circ}$ in radians, in terms of $\pi$ .

Show solutionHide solution

Colour keyfinal value

\displaystyle 150^{\circ} = 150 \times \dfrac{\pi}{180}

Multiply by

\dfrac{\pi}{180}

, the conversion factor from degrees to radians.

\displaystyle = \textcolor{#010106}{\dfrac{5\pi}{6}}

Cancel the common factor in the fraction.

Answer

150^{\circ} = \dfrac{5\pi}{6}

rad

Example 1.2

A negative angle

Express $-60^{\circ}$ in radians, in terms of $\pi$ .

Show solutionHide solution

Colour keyfinal value

\displaystyle -60^{\circ} = -60 \times \dfrac{\pi}{180}

Multiply by

\dfrac{\pi}{180}

, the conversion factor from degrees to radians.

\displaystyle = \textcolor{#010106}{-\dfrac{\pi}{3}}

Cancel the common factor in the fraction.

Answer

-60^{\circ} = -\dfrac{\pi}{3}

rad

Common angles in degrees and radians
Degrees	Radians
$0^{\circ}$	$0$
$30^{\circ}$	$\dfrac{\pi}{6}$
$45^{\circ}$	$\dfrac{\pi}{4}$
$60^{\circ}$	$\dfrac{\pi}{3}$
$90^{\circ}$	$\dfrac{\pi}{2}$
$180^{\circ}$	$\pi$
$270^{\circ}$	$\dfrac{3\pi}{2}$
$360^{\circ}$	$2\pi$

SGSchoolKaki Tip

Learn the radian values of $30^{\circ}$ , $45^{\circ}$ , $60^{\circ}$ and $90^{\circ}$ from the table: $\dfrac{\pi}{6}$ , $\dfrac{\pi}{4}$ , $\dfrac{\pi}{3}$ and $\dfrac{\pi}{2}$ . Every other multiple of these comes by adding them up.

Example 1.3

Radians to degrees

Express $\dfrac{7\pi}{6}$ radians in degrees.

Show solutionHide solution

Colour keyfinal value

\displaystyle \dfrac{7\pi}{6}\text{ rad}\;= \dfrac{7\pi}{6} \times \dfrac{180^{\circ}}{\pi}

Multiply by

\dfrac{180^{\circ}}{\pi}

, the conversion factor from radians to degrees.

\displaystyle = \textcolor{#010106}{210^{\circ}}

The

\pi

cancels, leaving a number of degrees.

Answer

\dfrac{7\pi}{6}

rad

= 210^{\circ}

Example 1.4

More than one full turn

Express $\dfrac{7\pi}{3}$ radians in degrees.

Show solutionHide solution

Colour keyfinal value

\displaystyle \dfrac{7\pi}{3}\text{ rad}\;= \dfrac{7\pi}{3} \times \dfrac{180^{\circ}}{\pi}

Multiply by

\dfrac{180^{\circ}}{\pi}

, the conversion factor from radians to degrees.

\displaystyle = \textcolor{#010106}{420^{\circ}}

The

\pi

cancels, leaving a number of degrees.

Answer

\dfrac{7\pi}{3}

rad

= 420^{\circ}

Example 1.5

A decimal number of radians

An angle measures $1.2$ radians. Use a calculator in radian mode to express it in degrees, correct to $1$ decimal place.

Show solutionHide solution

Colour keyfinal value

\displaystyle 1.2\text{ rad}\;= 1.2 \times \dfrac{180^{\circ}}{\pi}

Multiply by

\dfrac{180^{\circ}}{\pi}

to turn radians into degrees.

\displaystyle \approx \textcolor{#010106}{68.8^{\circ}}

Use the calculator and round to 1 decimal place.

Answer

1.2

rad

\approx 68.8^{\circ}

A calculator has a degree mode and a radian mode. Use radian mode for an angle in radians. In degree mode the calculator treats

1.2

1.2

degrees, not

1.2

radians, and every answer that follows is wrong.

The Core Idea

Arc length and sector area.

For a sector of radius $r$ and angle $\theta$ at the centre, arc length $s = r\theta$ and sector area $A = \dfrac{1}{2}r^{2}\theta$ . $\theta$ must be in radians. The sector is the fraction $\dfrac{\theta}{2\pi}$ of the whole circle, which is why both formulas come out so simply.

Example 1.6

Arc length and sector area

A sector of a circle has radius $6$ cm and angle $\dfrac{\pi}{3}$ radians at the centre. Find, in terms of $\pi$ , the length of the arc and the area of the sector.

Show solutionHide solution

Colour keyfinal value

\displaystyle s\;= r\theta = 6 \times \dfrac{\pi}{3} = \textcolor{#010106}{2\pi}

Arc length is radius times the angle, and the angle must be in radians.

\displaystyle A\;= \dfrac{1}{2} r^2 \theta = \dfrac{1}{2} \times 6^2 \times \dfrac{\pi}{3} = \textcolor{#010106}{6\pi}

Sector area is half the radius squared times the angle in radians.

Answer

arc length

= 2\pi

cm; area

= 6\pi

^2

Think About It

A semicircle has radius

5

cm. Try

s = r\theta

with

\theta = 180

and with

\theta = \pi

. Which result is the length of the semicircular arc?

Show the full workingHide the working

\displaystyle 5 \times 180\;= 900

With the angle written in degrees the 'arc' would come out far longer than the whole circle.

\displaystyle 5 \times \pi\;= 5\pi

With the angle in radians, a half turn is

\pi

\displaystyle \dfrac{1}{2} \times 2\pi \times 5\;= 5\pi

Half of the circumference of a circle of radius 5.

Answer

Only

5\pi

is the length of the semicircle: the formula

s = r\theta

gives the right answer only when

\theta

is in radians.

Example 1.7

A reflex sector

A sector of a circle has radius $8$ cm and angle $\dfrac{5\pi}{4}$ radians at the centre. Find, in terms of $\pi$ , the length of the arc and the area of the sector.

Show solutionHide solution

Colour keyfinal value

\displaystyle s\;= r\theta = 8 \times \dfrac{5\pi}{4} = \textcolor{#010106}{10\pi}

Arc length is radius times the angle, and the angle must be in radians.

\displaystyle A\;= \dfrac{1}{2} r^2 \theta = \dfrac{1}{2} \times 8^2 \times \dfrac{5\pi}{4} = \textcolor{#010106}{40\pi}

Sector area is half the radius squared times the angle in radians.

Answer

arc length

= 10\pi

cm; area

= 40\pi

^2

Supports K341-G1-01 · K341-G1-03

The six ratios and special angles

In a right-angled triangle the shape is fixed by one acute angle $\theta$ , and so are the ratios of the sides. Three ratios you already know are sine, cosine and tangent. Three more, the reciprocal ratios, turn those upside down.

SGSchoolKaki Tip

Draw the triangle first and label the sides opposite (opposite $\theta$ ), adjacent (next to $\theta$ and not the longest side) and hypotenuse (opposite the right angle) before you write any ratio.

The Core Idea

The six ratios.

$\sin\theta = \dfrac{\text{opp}}{\text{hyp}}$ , $\cos\theta = \dfrac{\text{adj}}{\text{hyp}}$ and $\tan\theta = \dfrac{\text{opp}}{\text{adj}}$ . The reciprocal ratios are $\sec\theta = \dfrac{1}{\cos\theta}$ , $\operatorname{cosec}\theta = \dfrac{1}{\sin\theta}$ and $\cot\theta = \dfrac{1}{\tan\theta}$ . Each one is undefined when the ratio it flips is zero.

Example 2.1

Three ratios from a triangle

A right-angled triangle has an acute angle $\theta$ . The side opposite $\theta$ has length $3$ and the side adjacent to $\theta$ has length $4$ . Find the length of the hypotenuse, and write down $\sin\theta$ , $\cos\theta$ and $\tan\theta$ as exact values.

Show solutionHide solution

Colour keybasic angle, side or quarter periodfinal value

\displaystyle \text{hyp}\;= \sqrt{3^2 + 4^2} = \sqrt{25} = \textcolor{#010105}{5}

Pythagoras: the hypotenuse squared is the sum of the squares of the other two sides.

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

Sin is opposite over hypotenuse.

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

Cos is adjacent over hypotenuse.

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

Tan is opposite over adjacent.

Answer

hypotenuse

= 5

;

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

;

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

;

\tan \theta = \dfrac{3}{4}

Your Turn 2.1

A right-angled triangle has an acute angle $\theta$ . The side opposite $\theta$ has length $5$ and the side adjacent to $\theta$ has length $12$ . Find the length of the hypotenuse, and write down $\sin\theta$ , $\cos\theta$ and $\tan\theta$ as exact values.

Show full solutionHide solution

Colour keybasic angle, side or quarter periodfinal value

\displaystyle \text{hyp}\;= \sqrt{5^2 + 12^2} = \sqrt{169} = \textcolor{#010105}{13}

Pythagoras: the hypotenuse squared is the sum of the squares of the other two sides.

\displaystyle \sin \theta\;= \textcolor{#010106}{\dfrac{5}{13}}

Sin is opposite over hypotenuse.

\displaystyle \cos \theta\;= \textcolor{#010106}{\dfrac{12}{13}}

Cos is adjacent over hypotenuse.

\displaystyle \tan \theta\;= \textcolor{#010106}{\dfrac{5}{12}}

Tan is opposite over adjacent.

Answer

hypotenuse

= 13

;

\sin \theta = \dfrac{5}{13}

;

\cos \theta = \dfrac{12}{13}

;

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

Example 2.2

The reciprocal ratios

A right-angled triangle has an acute angle $\theta$ . The side opposite $\theta$ has length $8$ and the side adjacent to $\theta$ has length $15$ . Find the length of the hypotenuse, and write down $\sec\theta$ , $\operatorname{cosec}\theta$ and $\cot\theta$ as exact values.

Show solutionHide solution

Colour keybasic angle, side or quarter periodfinal value

\displaystyle \text{hyp}\;= \sqrt{8^2 + 15^2} = \sqrt{289} = \textcolor{#010105}{17}

Pythagoras: the hypotenuse squared is the sum of the squares of the other two sides.

\displaystyle \sec \theta\;= \textcolor{#010106}{\dfrac{17}{15}}

Sec is hypotenuse over adjacent (the flip of cosine).

\displaystyle \operatorname{cosec} \theta\;= \textcolor{#010106}{\dfrac{17}{8}}

Cosec is hypotenuse over opposite (the flip of sine).

\displaystyle \cot \theta\;= \textcolor{#010106}{\dfrac{15}{8}}

Cot is adjacent over opposite (the flip of tangent).

Answer

hypotenuse

= 17

;

\sec \theta = \dfrac{17}{15}

;

\operatorname{cosec} \theta = \dfrac{17}{8}

;

\cot \theta = \dfrac{15}{8}

Your Turn 2.2

A right-angled triangle has an acute angle $\theta$ . The side opposite $\theta$ has length $20$ and the side adjacent to $\theta$ has length $21$ . Find the length of the hypotenuse, and write down $\sec\theta$ , $\operatorname{cosec}\theta$ and $\cot\theta$ as exact values.

Show full solutionHide solution

Colour keybasic angle, side or quarter periodfinal value

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

Pythagoras: the hypotenuse squared is the sum of the squares of the other two sides.

\displaystyle \sec \theta\;= \textcolor{#010106}{\dfrac{29}{21}}

Sec is hypotenuse over adjacent (the flip of cosine).

\displaystyle \operatorname{cosec} \theta\;= \textcolor{#010106}{\dfrac{29}{20}}

Cosec is hypotenuse over opposite (the flip of sine).

\displaystyle \cot \theta\;= \textcolor{#010106}{\dfrac{21}{20}}

Cot is adjacent over opposite (the flip of tangent).

Answer

hypotenuse

= 29

;

\sec \theta = \dfrac{29}{21}

;

\operatorname{cosec} \theta = \dfrac{29}{20}

;

\cot \theta = \dfrac{21}{20}

\operatorname{cosec}\theta

means

\dfrac{1}{\sin\theta}

. It is not the inverse function

\sin^{-1}

that a calculator key of that name gives, which is a different idea met in a later section of this chapter.

Example 2.3

A surd hypotenuse

A right-angled triangle has an acute angle $\theta$ . The side opposite $\theta$ has length $2$ and the side adjacent to $\theta$ has length $3$ . Find the length of the hypotenuse, and write down $\sin\theta$ , $\cos\theta$ and $\tan\theta$ as exact values.

Show solutionHide solution

Colour keybasic angle, side or quarter periodfinal value

\displaystyle \text{hyp}\;= \sqrt{2^2 + 3^2} = \textcolor{#010105}{\sqrt{13}}

Pythagoras: the hypotenuse squared is the sum of the squares of the other two sides.

\displaystyle \sin \theta\;= \dfrac{2}{\sqrt{13}} = \textcolor{#010106}{\dfrac{2 \sqrt{13}}{13}}

Sin is opposite over hypotenuse.

\displaystyle \cos \theta\;= \dfrac{3}{\sqrt{13}} = \textcolor{#010106}{\dfrac{3 \sqrt{13}}{13}}

Cos is adjacent over hypotenuse.

\displaystyle \tan \theta\;= \textcolor{#010106}{\dfrac{2}{3}}

Tan is opposite over adjacent.

Answer

hypotenuse

= \sqrt{13}

;

\sin \theta = \dfrac{2 \sqrt{13}}{13}

;

\cos \theta = \dfrac{3 \sqrt{13}}{13}

;

\tan \theta = \dfrac{2}{3}

Your Turn 2.3

A right-angled triangle has an acute angle $\theta$ . The side opposite $\theta$ has length $3$ and the side adjacent to $\theta$ has length $5$ . Find the length of the hypotenuse, and write down $\sin\theta$ , $\cos\theta$ and $\tan\theta$ as exact values.

Show full solutionHide solution

Colour keybasic angle, side or quarter periodfinal value

\displaystyle \text{hyp}\;= \sqrt{3^2 + 5^2} = \textcolor{#010105}{\sqrt{34}}

Pythagoras: the hypotenuse squared is the sum of the squares of the other two sides.

\displaystyle \sin \theta\;= \dfrac{3}{\sqrt{34}} = \textcolor{#010106}{\dfrac{3 \sqrt{34}}{34}}

Sin is opposite over hypotenuse.

\displaystyle \cos \theta\;= \dfrac{5}{\sqrt{34}} = \textcolor{#010106}{\dfrac{5 \sqrt{34}}{34}}

Cos is adjacent over hypotenuse.

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

Tan is opposite over adjacent.

Answer

hypotenuse

= \sqrt{34}

;

\sin \theta = \dfrac{3 \sqrt{34}}{34}

;

\cos \theta = \dfrac{5 \sqrt{34}}{34}

;

\tan \theta = \dfrac{3}{5}

The Core Idea

Exact values at $30^{\circ}$ , $45^{\circ}$ and $60^{\circ}$ .

Half of a square of side $1$ cut along its diagonal gives the $45^{\circ}$ triangle with sides $1$ , $1$ , $\sqrt{2}$ . Half of an equilateral triangle of side $2$ gives the $30^{\circ}$ and $60^{\circ}$ triangle with sides $1$ , $\sqrt{3}$ , $2$ . Reading ratios off these two triangles gives the values in the table, in degrees and in radians ( $30^{\circ} = \dfrac{\pi}{6}$ , $45^{\circ} = \dfrac{\pi}{4}$ , $60^{\circ} = \dfrac{\pi}{3}$ ).

Exact values at the three special angles
Angle	sin	cos	tan
$30^{\circ}$ or $\dfrac{\pi}{6}$	$\dfrac{1}{2}$	$\dfrac{\sqrt{3}}{2}$	$\dfrac{1}{\sqrt{3}}$
$45^{\circ}$ or $\dfrac{\pi}{4}$	$\dfrac{1}{\sqrt{2}}$	$\dfrac{1}{\sqrt{2}}$	$1$
$60^{\circ}$ or $\dfrac{\pi}{3}$	$\dfrac{\sqrt{3}}{2}$	$\dfrac{1}{2}$	$\sqrt{3}$

Example 2.4

A special angle in degrees

Use the special-angle values, with no calculator, to write $\sin 45^{\circ}$ exactly.

Show solutionHide solution

Colour keyfinal value

\displaystyle \text{hyp}\;= \sqrt{1^2 + 1^2} = \sqrt{2}

Half a unit square cut along its diagonal: both short sides are 1, so Pythagoras gives the hypotenuse.

\displaystyle \sin 45^{\circ}\;= \dfrac{1}{\sqrt{2}} = \textcolor{#010106}{\dfrac{\sqrt{2}}{2}}

Read the ratio off the triangle: opposite over hypotenuse.

Answer

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

Example 2.5

A special angle in radians

Use the special-angle values, with no calculator, to write $\tan \dfrac{\pi}{3}$ exactly.

Show solutionHide solution

Colour keyfinal value

\displaystyle \text{other side}\;= \sqrt{2^2 - 1^2} = \sqrt{3}

Half an equilateral triangle of side 2: the hypotenuse is 2, the short side is 1, so Pythagoras gives the third side.

\displaystyle \tan \dfrac{\pi}{3}\;= \dfrac{\sqrt{3}}{1} = \textcolor{#010106}{\sqrt{3}}

Read the ratio off the triangle: opposite over adjacent.

Answer

\tan \dfrac{\pi}{3} = \sqrt{3}

Think About It

Why is

\tan 45^{\circ}

exactly

1

? Use the triangle with two equal short sides.

Show the full workingHide the working

Let each of the two equal sides of the right-angled triangle have length

k

The

45^{\circ}

triangle is half of a square, so its two short sides are equal.

\displaystyle \tan 45^{\circ}\;= \dfrac{k}{k} = 1

Opposite and adjacent are the same length, so their ratio is 1.

Answer

The opposite and adjacent sides are equal, so

\tan 45^{\circ} = 1

whatever the size of the triangle.

Example 2.6

A reciprocal at a special angle

Find the exact value of $\sec 30^{\circ}$ .

Show solutionHide solution

Colour keyfinal value

\displaystyle \sec 30^{\circ}\;= \dfrac{1}{\cos 30^{\circ}}

Sec is the reciprocal of cos: find cos first.

\displaystyle \cos 30^{\circ}\;= \textcolor{#010106}{\dfrac{\sqrt{3}}{2}}

First quadrant: all ratios are positive, so read the special-angle value.

\displaystyle \sec 30^{\circ}\;= \dfrac{1}{\dfrac{\sqrt{3}}{2}} = \textcolor{#010106}{\dfrac{2 \sqrt{3}}{3}}

Flip the fraction, then rationalise the denominator if a surd is left.

Answer

\sec 30^{\circ} = \dfrac{2 \sqrt{3}}{3}

Your Turn 2.6

Find the exact value of $\operatorname{cosec} \dfrac{\pi}{4}$ .

Show full solutionHide solution

Colour keyfinal value

\displaystyle \operatorname{cosec} \dfrac{\pi}{4}\;= \dfrac{1}{\sin \dfrac{\pi}{4}}

Cosec is the reciprocal of sin: find sin first.

\displaystyle \sin \dfrac{\pi}{4}\;= \textcolor{#010106}{\dfrac{\sqrt{2}}{2}}

First quadrant: all ratios are positive, so read the special-angle value.

\displaystyle \operatorname{cosec} \dfrac{\pi}{4}\;= \dfrac{1}{\dfrac{\sqrt{2}}{2}} = \textcolor{#010106}{\sqrt{2}}

Flip the fraction, then rationalise the denominator if a surd is left.

Answer

\operatorname{cosec} \dfrac{\pi}{4} = \sqrt{2}

Supports K341-G1-01 · K341-G1-03

Angles of any size: quadrants and exact values

An angle does not have to be acute. Turn a line about the origin and its angle can be $150^{\circ}$ , $405^{\circ}$ or even negative for a clockwise turn. The ratios still make sense, and the quadrant the line lands in decides their signs.

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}}

, for an acute angle

\theta

. Every value below is read from such a triangle for the basic angle.

The Core Idea

Quadrant, basic angle, sign (ASTC).

Measure angles anticlockwise from the positive $x$ -axis. The basic angle $\alpha$ is the acute angle between the line and the $x$ -axis. The quadrant gives the sign: All positive in I, Sine in II, Tangent in III, Cosine in IV. Then fn(angle) = (sign) fn(basic angle). Reciprocal functions take the sign of the function they flip.

the four quadrants with the ratios that are positive in each

Example 3.1

Second quadrant

Find the exact value of $\sin 150^{\circ}$ .

Show solutionHide solution

Colour keybasic angle, side or quarter periodfinal value

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

In the second quadrant the basic angle is a half turn minus the angle.

\displaystyle \sin 150^{\circ}\;= \sin 30^{\circ} = \textcolor{#010106}{\dfrac{1}{2}}

In quadrant II only sine (and cosec) is positive, so the sign is positive.

Answer

\sin 150^{\circ} = \dfrac{1}{2}

Example 3.2

Third quadrant

Find the exact value of $\tan 225^{\circ}$ .

Show solutionHide solution

Colour keybasic angle, side or quarter periodfinal value

\displaystyle \alpha\;= 225^{\circ} - 180^{\circ} = \textcolor{#010105}{45^{\circ}}

In the third quadrant the basic angle is the angle minus a half turn.

\displaystyle \tan 225^{\circ}\;= \tan 45^{\circ} = \textcolor{#010106}{1}

In quadrant III only tangent (and cot) is positive, so the sign is positive.

Answer

\tan 225^{\circ} = 1

Example 3.3

Fourth quadrant

Find the exact value of $\cos 300^{\circ}$ .

Show solutionHide solution

Colour keybasic angle, side or quarter periodfinal value

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

In the fourth quadrant the basic angle is a full turn minus the angle.

\displaystyle \cos 300^{\circ}\;= \cos 60^{\circ} = \textcolor{#010106}{\dfrac{1}{2}}

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

Answer

\cos 300^{\circ} = \dfrac{1}{2}

The four quadrants: sign and basic angle
Quadrant	Angle from the positive x-axis	Positive ratios	Basic angle for an angle θ
I	$0^{\circ}$ to $90^{\circ}$	sin, cos and tan	$\theta$
II	$90^{\circ}$ to $180^{\circ}$	sin only	$180^{\circ} - \theta$
III	$180^{\circ}$ to $270^{\circ}$	tan only	$\theta - 180^{\circ}$
IV	$270^{\circ}$ to $360^{\circ}$	cos only	$360^{\circ} - \theta$

The Core Idea

Related angles.

Reading the table in terms of the angle $\theta$ gives the related-angle rules: $\sin(180^{\circ} - \theta) = \sin\theta$ , $\cos(180^{\circ} - \theta) = -\cos\theta$ , $\tan(180^{\circ} + \theta) = \tan\theta$ , $\sin(360^{\circ} - \theta) = -\sin\theta$ , $\cos(360^{\circ} - \theta) = \cos\theta$ , $\sin(-\theta) = -\sin\theta$ , $\cos(-\theta) = \cos\theta$ and $\tan(-\theta) = -\tan\theta$ . In radians replace $180^{\circ}$ by $\pi$ and $360^{\circ}$ by $2\pi$ .

Example 3.4

Beyond one full turn

Find the exact value of $\sin 405^{\circ}$ .

Show solutionHide solution

Colour keyfinal value

\displaystyle \sin 405^{\circ}\;= \sin 45^{\circ}

Remove whole turns: the ratio repeats every full turn.

\displaystyle = \textcolor{#010106}{\dfrac{\sqrt{2}}{2}}

First quadrant: all ratios are positive, so read the special-angle value.

Answer

\sin 405^{\circ} = \dfrac{\sqrt{2}}{2}

Example 3.5

A negative angle

Find the exact value of $\tan\left(-120^{\circ}\right)$ .

Show solutionHide solution

Colour keybasic angle, side or quarter periodfinal value

\displaystyle \tan\left(-120^{\circ}\right)\;= \tan 240^{\circ}

Add whole turns: the ratio repeats every full turn.

\displaystyle \alpha\;= 240^{\circ} - 180^{\circ} = \textcolor{#010105}{60^{\circ}}

In the third quadrant the basic angle is the angle minus a half turn.

\displaystyle = \tan 60^{\circ} = \textcolor{#010106}{\sqrt{3}}

In quadrant III only tangent (and cot) is positive, so the sign is positive.

Answer

\tan\left(-120^{\circ}\right) = \sqrt{3}

Your Turn 3.5

Find the exact value of $\cos\left(-210^{\circ}\right)$ .

Show full solutionHide solution

Colour keybasic angle, side or quarter periodfinal value

\displaystyle \cos\left(-210^{\circ}\right)\;= \cos 150^{\circ}

Add whole turns: the ratio repeats every full turn.

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

In the second quadrant the basic angle is a half turn minus the angle.

\displaystyle = -\cos 30^{\circ} = \textcolor{#010106}{- \dfrac{\sqrt{3}}{2}}

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

Answer

\cos\left(-210^{\circ}\right) = - \dfrac{\sqrt{3}}{2}

SGSchoolKaki Tip

For any angle: (1) take away whole turns, or add them for a negative angle, until the angle is between $0^{\circ}$ and $360^{\circ}$ ; (2) name the quadrant; (3) find the basic angle; (4) choose the sign from ASTC; (5) read the special-angle value.

the same four quadrants with angles in radians

Example 3.6

Radians in the second and third quadrants

Find the exact value of $\tan \dfrac{2\pi}{3}$ .

Show solutionHide solution

Colour keybasic angle, side or quarter periodfinal value

\displaystyle \alpha\;= \pi - \dfrac{2\pi}{3} = \textcolor{#010105}{\dfrac{\pi}{3}}

In the second quadrant the basic angle is a half turn minus the angle.

\displaystyle \tan \dfrac{2\pi}{3}\;= -\tan \dfrac{\pi}{3} = \textcolor{#010106}{- \sqrt{3}}

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

Answer

\tan \dfrac{2\pi}{3} = - \sqrt{3}

Example 3.7

A reciprocal function in radians

Find the exact value of $\operatorname{cosec} \dfrac{7\pi}{6}$ .

Show solutionHide solution

Colour keybasic angle, side or quarter periodfinal value

\displaystyle \operatorname{cosec} \dfrac{7\pi}{6}\;= \dfrac{1}{\sin \dfrac{7\pi}{6}}

Cosec is the reciprocal of sin: find sin first.

\displaystyle \alpha\;= \dfrac{7\pi}{6} - \pi = \textcolor{#010105}{\dfrac{\pi}{6}}

In the third quadrant the basic angle is the angle minus a half turn.

\displaystyle \sin \dfrac{7\pi}{6}\;= -\sin \dfrac{\pi}{6} = \textcolor{#010106}{- \dfrac{1}{2}}

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

\displaystyle \operatorname{cosec} \dfrac{7\pi}{6}\;= \dfrac{1}{- \dfrac{1}{2}} = \textcolor{#010106}{-2}

Flip the fraction, then rationalise the denominator if a surd is left.

Answer

\operatorname{cosec} \dfrac{7\pi}{6} = -2

Your Turn 3.7

Find the exact value of $\sec \dfrac{5\pi}{4}$ .

Show full solutionHide solution

Colour keybasic angle, side or quarter periodfinal value

\displaystyle \sec \dfrac{5\pi}{4}\;= \dfrac{1}{\cos \dfrac{5\pi}{4}}

Sec is the reciprocal of cos: find cos first.

\displaystyle \alpha\;= \dfrac{5\pi}{4} - \pi = \textcolor{#010105}{\dfrac{\pi}{4}}

In the third quadrant the basic angle is the angle minus a half turn.

\displaystyle \cos \dfrac{5\pi}{4}\;= -\cos \dfrac{\pi}{4} = \textcolor{#010106}{- \dfrac{\sqrt{2}}{2}}

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

\displaystyle \sec \dfrac{5\pi}{4}\;= \dfrac{1}{- \dfrac{\sqrt{2}}{2}} = \textcolor{#010106}{- \sqrt{2}}

Flip the fraction, then rationalise the denominator if a surd is left.

Answer

\sec \dfrac{5\pi}{4} = - \sqrt{2}

\tan 90^{\circ}

is undefined, and so are

\sec 90^{\circ}

\operatorname{cosec} 0^{\circ}

and

\cot 0^{\circ}

: each would need a division by zero. A calculator shows an error for them, and no exam question asks for them.

Example 3.8

A quadrantal angle

Write down the value of $\sin 270^{\circ}$ .

Show solutionHide solution

Colour keyfinal value

\displaystyle \sin 270^{\circ}\;= \textcolor{#010106}{-1}

On the unit circle this angle is the point

(0, -1)

: sine is its

y

-coordinate, cosine its

x

-coordinate, tangent is

y

over

x

Answer

\sin 270^{\circ} = -1

Think About It

Why is

\sin 150^{\circ}

equal to

\sin 30^{\circ}

? Compare the two angles with the horizontal axis.

Show the full workingHide the working

\displaystyle 180^{\circ} - 150^{\circ}\;= 30^{\circ}

The basic angle: how far

150^{\circ}

is from the horizontal axis.

\displaystyle \sin 150^{\circ}\;= \sin 30^{\circ} = \dfrac{1}{2}

Sine is positive in the second quadrant, so no sign change.

Answer

Both angles make the same acute angle with the horizontal axis and sine is positive in the second quadrant, so the two sines are equal.

Example 3.9

A ratio and a quadrant given

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

Show solutionHide solution

Colour keybasic angle, side or quarter period

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 3.9

Given that $\sin A = \dfrac{7}{25}$ and that $A$ lies in the second quadrant, find the exact values of $\cos A$ and $\tan A$ .

Show full solutionHide solution

Colour keybasic angle, side or quarter period

The basic-angle triangle has opposite 7 and hypotenuse 25.

\displaystyle \text{adj}\;= \sqrt{25^2 - 7^2} = \sqrt{576} = \textcolor{#010105}{24}

Pythagoras gives the third side of the triangle.

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

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

\displaystyle \tan A\;= -\dfrac{7}{24}

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

Answer

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

;

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

Example 3.10

A tangent and a quadrant given

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

Show solutionHide solution

Colour keybasic angle, side or quarter periodfinal value

The basic-angle triangle has opposite 3 and adjacent 4.

\displaystyle \text{hyp}\;= \sqrt{3^2 + 4^2} = \sqrt{25} = \textcolor{#010105}{5}

Pythagoras gives the third side of the triangle.

\displaystyle \sin A\;= -\dfrac{3}{5}

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

\displaystyle \cos A\;= \textcolor{#010106}{\dfrac{4}{5}}

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

Answer

\sin A = - \dfrac{3}{5}

;

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

Your Turn 3.10

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

Show full solutionHide solution

Colour keybasic angle, side or quarter period

The basic-angle triangle has opposite 5 and adjacent 12.

\displaystyle \text{hyp}\;= \sqrt{5^2 + 12^2} = \sqrt{169} = \textcolor{#010105}{13}

Pythagoras gives the third side of the triangle.

\displaystyle \sin A\;= -\dfrac{5}{13}

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

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

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

Answer

\sin A = - \dfrac{5}{13}

;

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

Links To

Equations such as $\sin x = \dfrac{1}{2}$ for all angles in an interval build on the quadrant rules above, in Chapter 10: Trigonometric identities, equations and proofs

Locked · Chapter 9

You've read the free 28%. Unlock the rest of Chapter 9.

Still to read in this chapter:

Principal values
Basic graphs
a, b and c
Tan and constants
Formula panel
Exercise

$9 one-time · all 17 chapters · no subscription

Worked examples, formulas and exam checks.
The whole A-Math syllabus.

ONE PRICE17chapters

Premium members get all 17 chapters included.

Notes version 1.0 · Last updated 3 Nov 2026 · Chapter 9 of 17