Trig Ratios of Any Angle

Find sin, cos and tan of any angle in degrees or radians, exactly where possible, using quadrant signs and the related acute angle; plus sec, cosec and cot for 0606/4037.

In the exam for IGCSE Additional Maths 0606, Edexcel Further Pure Maths 4PM1 · 7 real past-paper questions below

اقرأ هذا الدرس مشروحاً بالعربي ←

Beyond the triangle

In a right-angled triangle every angle is less than 90°. But a wheel can turn 200°, and a clock hand goes past 360°. To use trig on any angle, we swap the triangle for a circle.

Draw a circle of radius 1 with its centre at the origin O. Start at the point $(1, 0)$ and turn anticlockwise through an angle $θ$ . You land on a point P on the circle.

Rule

The x-coordinate of P is $cos θ$ . The y-coordinate of P is $sin θ$ . This is the meaning of sin and cos for every angle, not only acute ones.

Why does this agree with SOH CAH TOA? For an acute angle, drop a line from P down to the x-axis. The hypotenuse is the radius, 1. So the opposite side is $sin θ$ and the adjacent side is $cos θ$ .

Tan was opposite over adjacent. On the circle that is the height of P over its across distance:

Formula

tan θ = \frac{s i n θ}{c o s θ}

$sin θ$ — the y-coordinate of P: how far up (or down) it is.
$cos θ$ — the x-coordinate of P: how far right (or left) it is.
$θ$ — the angle turned from the positive x-axis.

Worked example

1
Use the circle to find $sin$ , $cos$ and $tan$ of $18 0^{\circ}$ and of $9 0^{\circ}$ .
2
$P = (- 1, 0)$ — Half a turn from (1, 0) takes you to the far left of the circle.
3
$sin 18 0^{\circ} = 0, cos 18 0^{\circ} = - 1$ — Read the y-coordinate and the x-coordinate.
4
$tan 18 0^{\circ} = \frac{0}{- 1} = 0$ — Use tan = sin over cos.
5
$9 0^{\circ} : P = (0, 1)$ — A quarter turn takes you to the top of the circle.
6
$sin 9 0^{\circ} = 1, cos 9 0^{\circ} = 0$ — Read the two coordinates.
7
$tan 9 0^{\circ} = \frac{1}{0}$ — You cannot divide by 0, so tan 90° does not exist.

Rule

Turn anticlockwise for a positive angle and clockwise for a negative angle. After a full turn ( $36 0^{\circ}$ or $2 π$ ) you go round again, so $40 0^{\circ}$ lands on the same point as $4 0^{\circ}$ .

Rule

4PM1 prints $tan θ = \frac{s i n θ}{c o s θ}$ on the formula sheet. 0606 and 4037 do not, so learn it.

Key idea

The word sine has a winding history. Indian mathematicians called it jya, a bowstring. Arabic scholars wrote it as jiba, and since Arabic is written without short vowels, it was later read as jaib, a fold or pocket. Latin translators in the 1100s turned that into sinus, and we got sine.

Signs in four quadrants

The two axes cut the circle into four quarters, called quadrants. We number them anticlockwise, starting top right. In each one, the signs of x and y are fixed, so the signs of the ratios are fixed too.

First quadrant, $0^{\circ}$ to $9 0^{\circ}$ : x and y are both positive. All three ratios are positive.
Second quadrant, $9 0^{\circ}$ to $18 0^{\circ}$ : y is positive, x is negative. Only sin is positive.
Third quadrant, $18 0^{\circ}$ to $27 0^{\circ}$ : both are negative, so y over x is positive. Only tan is positive.
Fourth quadrant, $27 0^{\circ}$ to $36 0^{\circ}$ : x is positive, y is negative. Only cos is positive.

Memory trick

CAST: write C, A, S, T in the quadrants starting bottom right and going anticlockwise. The letter in a quadrant tells you which ratio is positive there (A means all).

Every angle also has a related acute angle: the acute angle between the line OP and the x-axis. The ratio of the big angle has the same size as the ratio of the related angle. Only the sign can change.

Steps

1
1. Find the quadrant the angle is in.
2. Find the related acute angle $α$ between OP and the x-axis.
3. Write the same ratio of $α$ .
4. Put a + or − sign in front, using CAST.

Worked example

1
Write $cos 25 0^{\circ}$ as a ratio of an acute angle.
2
$18 0^{\circ} < 25 0^{\circ} < 27 0^{\circ}$ — 250° is in the third quadrant.
3
$α = 25 0^{\circ} - 18 0^{\circ} = 7 0^{\circ}$ — The line OP is 70° past the negative x-axis.
4
$cos 25 0^{\circ} = - cos 7 0^{\circ}$ — Only tan is positive in the third quadrant, so cos is negative.

Worked example

1
Write $sin (- 4 0^{\circ})$ as a ratio of an acute angle.
2
$- 4 0^{\circ}$ — Negative, so turn 40° clockwise. You land in the fourth quadrant.
3
$α = 4 0^{\circ}$ — OP is 40° below the positive x-axis.
4
$sin (- 4 0^{\circ}) = - sin 4 0^{\circ}$ — Only cos is positive in the fourth quadrant, so sin is negative.

One pair is worth learning by heart. The angles $θ$ and $18 0^{\circ} - θ$ are mirror images in the y-axis, so they have the same height:

Formula

sin (18 0^{\circ} - θ) = sin θ

$θ$ — any angle. In radians, $18 0^{\circ}$ becomes $π$ . No board prints this: learn it, or see it on the circle.

Worked example

1
Find another angle between $0^{\circ}$ and $18 0^{\circ}$ with the same sine as $3 5^{\circ}$ .
2
$18 0^{\circ} - 3 5^{\circ} = 14 5^{\circ}$ — Mirror the angle in the y-axis.
3
$sin 14 5^{\circ} = sin 3 5^{\circ}$ — Same height on the circle, so the same sine.

Common mistake

Using the angle between OP and the y-axis. The related angle is always measured to the x-axis: for $25 0^{\circ}$ it is $7 0^{\circ}$ , not $2 0^{\circ}$ .

Key idea

The related acute angle gives the size. The quadrant gives the sign.

Exact values

You already know the exact ratios of $3 0^{\circ}$ , $4 5^{\circ}$ and $6 0^{\circ}$ from half a square and half an equilateral triangle. Here they are again, with the radian names.

$θ$	$3 0^{\circ} = \frac{π}{6}$	$4 5^{\circ} = \frac{π}{4}$	$6 0^{\circ} = \frac{π}{3}$
$sin$	$\frac{1}{2}$	$\frac{2}{2}$	$\frac{3}{2}$
$cos$	$\frac{3}{2}$	$\frac{2}{2}$	$\frac{1}{2}$
$tan$	$\frac{3}{3}$	$1$	$3$

Rule

Learn this table. No board prints it. 0606 Paper 1 has no calculator, so these values are how you give exact answers there.

Now put them together with the four steps. The related angle gives the number from the table, and CAST gives the sign.

Worked example

1
Find the exact value of $sin 24 0^{\circ}$ .
2
$18 0^{\circ} < 24 0^{\circ} < 27 0^{\circ}$ — Third quadrant.
3
$α = 24 0^{\circ} - 18 0^{\circ} = 6 0^{\circ}$ — The related acute angle.
4
$sin 24 0^{\circ} = - sin 6 0^{\circ}$ — Only tan is positive in the third quadrant.
5
$= - \frac{3}{2}$ — Read sin 60° from the table.

Worked example

1
Find the exact value of $cos \frac{3 π}{4}$ .
2
$\frac{π}{2} < \frac{3 π}{4} < π$ — Between a quarter turn and a half turn: second quadrant.
3
$α = π - \frac{3 π}{4} = \frac{π}{4}$ — In radians, a half turn is π.
4
$cos \frac{3 π}{4} = - cos \frac{π}{4}$ — Only sin is positive in the second quadrant.
5
$= - \frac{2}{2}$ — Read cos 45° from the table.

Worked example

1
Find the exact value of $tan 33 0^{\circ}$ .
2
$27 0^{\circ} < 33 0^{\circ} < 36 0^{\circ}$ — Fourth quadrant.
3
$α = 36 0^{\circ} - 33 0^{\circ} = 3 0^{\circ}$ — 30° short of a full turn.
4
$tan 33 0^{\circ} = - tan 3 0^{\circ} = - \frac{3}{3}$ — Only cos is positive in the fourth quadrant, so tan is negative.

Common mistake

Getting the size right but the sign wrong. $sin 24 0^{\circ}$ is $- \frac{3}{2}$ , not $\frac{3}{2}$ . Always ask which quadrant before you write the answer.

Common mistake

Calculator in the wrong mode. If the angle is in radians, set the calculator to radians first, or every answer will be wrong.

Key idea

Table for the number, CAST for the sign. With these two, you never need a calculator for multiples of $3 0^{\circ}$ and $4 5^{\circ}$ .

sec, cosec and cot

0606 and 4037 use three more functions. 4PM1 does not use them, so 4PM1 students can skip this step. Each one is simply 1 over a ratio you already know.

Formula

s e c θ = \frac{1}{c o s θ}

$sec θ$ — secant: 1 over the cosine of the same angle.

Formula

cosec θ = \frac{1}{s i n θ}

$cosec θ$ — cosecant: 1 over the sine of the same angle.

Formula

c o t θ = \frac{1}{t a n θ} = \frac{cos θ}{sin θ}

$cot θ$ — cotangent: 1 over the tangent, which is cos over sin.

Memory trick

Look at the third letter. seC goes with Cos, coSec goes with Sin, coT goes with Tan.

Rule

Learn these three. They are not in the List of formulas. Each one has the same sign as its partner: sec is negative wherever cos is negative.

Worked example

1
Find the exact value of $sec 15 0^{\circ} + cot 22 5^{\circ}$ .
2
$cos 15 0^{\circ} = - cos 3 0^{\circ} = - \frac{3}{2}$ — Second quadrant, related angle 30°, cos negative.
3
$sec 15 0^{\circ} = \frac{1}{- \frac{3}{2}} = - \frac{2}{3} = - \frac{2 3}{3}$ — Flip the fraction, then rationalise the denominator.
4
$tan 22 5^{\circ} = tan 4 5^{\circ} = 1$ — Third quadrant, related angle 45°, tan positive.
5
$cot 22 5^{\circ} = \frac{1}{1} = 1$ — 1 over the tan.
6
$sec 15 0^{\circ} + cot 22 5^{\circ} = 1 - \frac{2 3}{3}$ $= \frac{3 - 2 3}{3}$ — Add, over a common denominator of 3.

Worked example

1
Find the exact value of $cosec \frac{7 π}{6}$ .
2
$π < \frac{7 π}{6} < \frac{3 π}{2}$ — Third quadrant.
3
$sin \frac{7 π}{6} = - sin \frac{π}{6} = - \frac{1}{2}$ — Related angle π/6, and sin is negative here.
4
$cosec \frac{7 π}{6} = \frac{1}{- \frac{1}{2}} = - 2$ — 1 over the sine.

Common mistake

Thinking $sec θ$ is $cos^{- 1} θ$ . The calculator key $cos^{- 1}$ gives an angle. $sec θ$ is a number: 1 divided by $cos θ$ .

Key idea

Find sin, cos or tan first. Then turn it upside down.

From one ratio to the rest

A common question gives you one ratio and tells you which quadrant the angle is in. Then it asks for the other ratios, exactly. A right-angled triangle gives the size; the quadrant gives the sign.

Steps

1
1. Ignore any minus sign. Draw a right-angled triangle that matches the ratio you are given.
2. Use Pythagoras to find the third side.
3. Read off the ratio you need.
4. Put the sign that matches the quadrant.

Worked example

1
Given that $tan θ = \frac{8}{15}$ and $18 0^{\circ} < θ < 27 0^{\circ}$ , find the exact values of $sin θ$ and $cos θ$ .
2
$opposite = 8, adjacent = 15$ — Tan is opposite over adjacent.
3
$hypotenuse = 8^{2} + 1 5^{2} = 289 = 17$ — Pythagoras.
4
$size of sin θ = \frac{8}{17}, size of cos θ = \frac{15}{17}$ — Opposite over hypotenuse, adjacent over hypotenuse.
5
$sin θ = - \frac{8}{17}, cos θ = - \frac{15}{17}$ — Third quadrant: only tan is positive, so sin and cos are both negative.

For 0606 and 4037, the question may go on to ask for sec, cosec or cot. Use the values you have just found.

Worked example

1
With the same $θ$ , find the exact value of $cosec θ + cot θ$ .
2
$cosec θ = \frac{1}{s i n θ} = - \frac{17}{8}$ — Flip sin θ. The sign stays negative.
3
$cot θ = \frac{1}{t a n θ} = \frac{15}{8}$ — Flip tan θ, which is positive.
4
$- \frac{17}{8} + \frac{15}{8} = - \frac{2}{8} = - \frac{1}{4}$ — Same denominator, so add the tops.

Sometimes the ratio is a letter, not a number. The method is the same: Pythagoras works with letters too.

Worked example

1
Given that $cos x = k$ , where $\frac{3 π}{2} < x < 2 π$ , find $sin x$ and $tan x$ in terms of $k$ .
2
$adjacent = k, hypotenuse = 1$ — Cos is adjacent over hypotenuse, and k is k over 1.
3
$opposite = 1 - k^{2}$ — Pythagoras: hypotenuse squared minus adjacent squared.
4
$sin x = - 1 - k^{2}$ — Fourth quadrant: only cos is positive, so sin is negative.
5
$tan x = - \frac{1 - k ^{2}}{k}$ — Tan is sin over cos.

Common mistake

Leaving out the sign. Mark schemes give the final mark only with the correct sign, so a positive $sin θ$ in the third quadrant loses it.

Common mistake

Using a calculator to find $θ$ first and then rounding. The question wants exact values, so use the triangle.

Key idea

The triangle gives you the numbers. The quadrant gives you the signs. Do them in that order.

Which method? Read the question

The wording of the question tells you which tool to pick up.

You see	You do
“Find the exact value of $sin 30 0^{\circ}$ ”	Related acute angle, exact value from the table, sign from CAST
An angle in radians, like $\frac{5 π}{4}$	Same method. A half turn is $π$ , a full turn is $2 π$
A negative angle	Turn clockwise, then find the quadrant
“Given $tan θ$ = … and $θ$ is obtuse”	Right-angled triangle and Pythagoras, then sign from the quadrant
“in terms of $k$ ”	Triangle with letters, then sign from the quadrant
$sec$ , $cosec$ , $cot$ (0606 and 4037)	Find cos, sin or tan, then take 1 over it

Common mistake

Answering with a rounded decimal when the question says exact. On 0606 Paper 1 you have no calculator, and an answer like $- 0.866$ gets no credit where $- \frac{3}{2}$ is asked for.

Key idea

Every angle lands on the same circle. Find where it lands, read the size from an acute angle, and let the quadrant decide the sign.

Practise trig ratios of any angle now

Interactive visual aids, then real past-paper questions marked on your working like the exam. Free to start.

Practise free on Lemmara

Real past-paper questions on trig ratios of any angle

Each with a full worked solution.

Lemmara is an independent study tool, not affiliated with or endorsed by Cambridge International Education or Pearson Edexcel. The lessons and worked examples are Lemmara’s own.