Home / IGCSE Add Maths 0606 / Trigonometry / The six trigonometric functions
Cambridge IGCSE · Topic 10.1
IGCSE Add Maths 0606 / Topic 10 / IGCSE

Six ratios, any angle, exact values.

Sine, cosine and tangent each have a reciprocal: secant, cosecant and cotangent. The unit circle extends all six to angles of any size, positive or negative, and the CAST diagram tells you the sign in each quadrant. With the exact values for 30, 45 and 60 degrees, one ratio and a quadrant are enough to find the other five without a calculator.

Full-screen labTrig graphs lab → Topic 10All trigonometry lessons →
The key idea

\(\sec\theta = \dfrac{1}{\cos\theta}\), \(\operatorname{cosec}\theta = \dfrac{1}{\sin\theta}\) and \(\cot\theta = \dfrac{1}{\tan\theta} = \dfrac{\cos\theta}{\sin\theta}\). For an angle of any size, the point at angle \(\theta\) on the unit circle is \((\cos\theta, \sin\theta)\). The size of any ratio comes from the reference angle (the acute angle to the \(x\)-axis); the sign comes from the quadrant: All positive in the first, Sine in the second, Tangent in the third, Cosine in the fourth (CAST). Each reciprocal has the same sign as its partner.

θ 180° − θ P(cos θ, sin θ) O 1 x y −1 1 A S T C Positive ratios A 1st: all six S 2nd: sin, cosec T 3rd: tan, cot C 4th: cos, sec reference angle: acute angle to the x-axis
Fig. 1: the point P at angle θ on the unit circle has coordinates (cos θ, sin θ). Here θ is in the second quadrant, so its reference angle is 180° - θ, and only sin and cosec are positive there. The CAST letters show which ratios are positive in each quadrant
Section 01

Three new reciprocal functions.

You already know sine, cosine and tangent. Each one has a reciprocal with its own name. A quick way to pair them: look at the third letter. sec goes with cos, cosec goes with sin, cot goes with tan. On Paper 1 there is no calculator, so you need the exact values for \(30^{\circ}\), \(45^{\circ}\) and \(60^{\circ}\) (in radians \(\frac{\pi}{6}\), \(\frac{\pi}{4}\), \(\frac{\pi}{3}\)); the reciprocals follow by flipping.

\[\sec\theta = \frac{1}{\cos\theta}, \qquad \operatorname{cosec}\theta = \frac{1}{\sin\theta}, \qquad \cot\theta = \frac{1}{\tan\theta} = \frac{\cos\theta}{\sin\theta}\]
Worked example Reciprocal exact values
Find the exact values of \(\sec 30^{\circ}\), \(\operatorname{cosec} 45^{\circ}\) and \(\cot 60^{\circ}\), rationalising any denominators.
  1. \(\cos 30^{\circ} = \dfrac{\sqrt{3}}{2}\), so \(\sec 30^{\circ} = \dfrac{2}{\sqrt{3}} = \dfrac{2\sqrt{3}}{3}\).
  2. \(\sin 45^{\circ} = \dfrac{1}{\sqrt{2}}\), so \(\operatorname{cosec} 45^{\circ} = \sqrt{2}\).
  3. \(\tan 60^{\circ} = \sqrt{3}\), so \(\cot 60^{\circ} = \dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}\).
Answer: \(\sec 30^{\circ} = \dfrac{2\sqrt{3}}{3}\), \(\operatorname{cosec} 45^{\circ} = \sqrt{2}\), \(\cot 60^{\circ} = \dfrac{\sqrt{3}}{3}\)
Worked example An expression in radians
Without a calculator, find the exact value of \(\operatorname{cosec}^{2}\dfrac{\pi}{6} + \cot\dfrac{\pi}{4}\sec\dfrac{\pi}{3}\).
  1. \(\sin\dfrac{\pi}{6} = \dfrac{1}{2}\), so \(\operatorname{cosec}\dfrac{\pi}{6} = 2\) and \(\operatorname{cosec}^{2}\dfrac{\pi}{6} = 4\).
  2. \(\tan\dfrac{\pi}{4} = 1\), so \(\cot\dfrac{\pi}{4} = 1\).
  3. \(\cos\dfrac{\pi}{3} = \dfrac{1}{2}\), so \(\sec\dfrac{\pi}{3} = 2\).
  4. \(4 + 1 \times 2 = 6\).
Answer: \(6\)
Your turn
Find the exact value of \(\cot 30^{\circ} \times \sec 45^{\circ}\).
Examiner trap

\(\sec\theta\) means \(\dfrac{1}{\cos\theta}\), not \(\cos^{-1}\theta\). The inverse function \(\cos^{-1}\) gives an angle; \(\sec\) gives a number. And do not pair sec with sin just because both start with s: use the third letter.

Section 02

Angles of any size.

Measure \(\theta\) anticlockwise from the positive \(x\)-axis (clockwise for a negative angle). The point at angle \(\theta\) on a circle of radius 1 is \((\cos\theta, \sin\theta)\), which defines sine and cosine for every angle, and \(\tan\theta = \dfrac{\sin\theta}{\cos\theta}\). Adding or subtracting \(360^{\circ}\) (or \(2\pi\)) lands on the same point, so it changes nothing. To evaluate any ratio: find the quadrant, find the reference angle (the acute angle between the radius and the \(x\)-axis, Fig. 1), use the exact value for that, then attach the sign from CAST.

Worked example A third-quadrant angle
Find the exact values of \(\sin 240^{\circ}\) and \(\sec 240^{\circ}\).
  1. \(240^{\circ}\) is in the third quadrant; the reference angle is \(240^{\circ} - 180^{\circ} = 60^{\circ}\).
  2. In the third quadrant only tan and cot are positive, so sin and cos are negative.
  3. \(\sin 240^{\circ} = -\sin 60^{\circ} = -\dfrac{\sqrt{3}}{2}\).
  4. \(\cos 240^{\circ} = -\cos 60^{\circ} = -\dfrac{1}{2}\), so \(\sec 240^{\circ} = -2\).
Answer: \(\sin 240^{\circ} = -\dfrac{\sqrt{3}}{2}\), \(\sec 240^{\circ} = -2\)
Worked example Negative and large angles
Find the exact values of (a) \(\cot\left(-\dfrac{5\pi}{6}\right)\) (b) \(\operatorname{cosec} 495^{\circ}\).
  1. (a) Add \(2\pi\): \(-\dfrac{5\pi}{6} + 2\pi = \dfrac{7\pi}{6}\), in the third quadrant. Reference angle \(\dfrac{7\pi}{6} - \pi = \dfrac{\pi}{6}\).
  2. Tan is positive in the third quadrant, so cot is too: \(\cot\left(-\dfrac{5\pi}{6}\right) = \cot\dfrac{\pi}{6} = \sqrt{3}\).
  3. (b) Subtract \(360^{\circ}\): \(495^{\circ} - 360^{\circ} = 135^{\circ}\), in the second quadrant. Reference angle \(45^{\circ}\).
  4. Sin is positive in the second quadrant: \(\operatorname{cosec} 495^{\circ} = \operatorname{cosec} 45^{\circ} = \sqrt{2}\).
Answer: (a) \(\sqrt{3}\) (b) \(\sqrt{2}\)
Your turn
Find the exact value of \(\sec(-135^{\circ})\).
Examiner trap

The reference angle is measured to the \(x\)-axis, never to the \(y\)-axis. For \(120^{\circ}\) it is \(180^{\circ} - 120^{\circ} = 60^{\circ}\), not \(120^{\circ} - 90^{\circ} = 30^{\circ}\); using \(30^{\circ}\) swaps the sine and cosine values.

Stage 1 · Learn

Check your understanding

Four quick checks. Each correct answer earns XP and records this skill on your profile.

Quick check+10 XP

What is the exact value of \(\sec 60^{\circ}\)?

Quick check+10 XP

In which quadrant is \(\sin\theta\) negative but \(\cos\theta\) positive?

Quick check+10 XP

What is the reference angle for \(230^{\circ}\)?

Quick check+10 XP

What is the exact value of \(\operatorname{cosec} 210^{\circ}\)?

Section 03

One ratio gives all six.

If you know one ratio and the quadrant, you can find the other five exactly. Ignore signs at first: draw a right-angled triangle with the reference angle, put in the two sides the ratio gives and find the third side by Pythagoras. Read off the sizes of the ratios from the triangle, then give each one the sign that CAST gives for that quadrant.

Worked example Tangent given, angle obtuse
Given that \(\tan A = -\dfrac{3}{4}\) and \(A\) is obtuse, find the exact values of \(\sec A\) and \(\operatorname{cosec} A\).
  1. \(A\) is obtuse, so it is in the second quadrant, where sin and cosec are positive and cos, sec are negative.
  2. Triangle: opposite 3, adjacent 4, so the hypotenuse is \(\sqrt{3^{2} + 4^{2}} = 5\).
  3. Sizes: \(\cos = \dfrac{4}{5}\), \(\sin = \dfrac{3}{5}\). Signs: \(\cos A = -\dfrac{4}{5}\), \(\sin A = \dfrac{3}{5}\).
  4. \(\sec A = -\dfrac{5}{4}\), \(\operatorname{cosec} A = \dfrac{5}{3}\).
Answer: \(\sec A = -\dfrac{5}{4}\), \(\operatorname{cosec} A = \dfrac{5}{3}\)
Worked example Secant given, fourth quadrant
Given that \(\sec x = 3\) and \(\dfrac{3\pi}{2} < x < 2\pi\), find the exact values of \(\tan x\), \(\cot x\) and \(\operatorname{cosec} x\).
  1. \(\cos x = \dfrac{1}{3}\). Triangle: adjacent 1, hypotenuse 3, so the opposite side is \(\sqrt{9 - 1} = 2\sqrt{2}\).
  2. \(x\) is in the fourth quadrant: only cos and sec are positive, so sin, tan, cosec and cot are negative.
  3. \(\tan x = -2\sqrt{2}\), so \(\cot x = -\dfrac{1}{2\sqrt{2}} = -\dfrac{\sqrt{2}}{4}\).
  4. \(\sin x = -\dfrac{2\sqrt{2}}{3}\), so \(\operatorname{cosec} x = -\dfrac{3}{2\sqrt{2}} = -\dfrac{3\sqrt{2}}{4}\).
Answer: \(\tan x = -2\sqrt{2}\), \(\cot x = -\dfrac{\sqrt{2}}{4}\), \(\operatorname{cosec} x = -\dfrac{3\sqrt{2}}{4}\)
Your turn
Given that \(\sin\theta = \dfrac{5}{13}\) and \(90^{\circ} < \theta < 180^{\circ}\), find \(\cos\theta\) and \(\cot\theta\).
Examiner trap

Keep the triangle sides positive and add signs only at the end. Putting \(-3\) on the triangle for \(\tan A = -\frac{3}{4}\) leads to a negative sine in the second quadrant, which is wrong. The hypotenuse (the radius) is always positive.

Stage 2 · Exam

Exam-style questions

Unlocks once the four checks above are done. Worth more XP, written to IGCSE Additional Mathematics Paper 1 and Paper 2 standard.

Finish the four checks above to unlock the exam questions
Exam style+20 XP

Find the exact value of \(\cot 300^{\circ}\).

Exam style+20 XP

Given that \(\tan A = -\dfrac{5}{12}\) and \(90^{\circ} < A < 180^{\circ}\), find the exact value of \(\operatorname{cosec} A\).

Exam style+20 XP

Find the exact value of \(\sec\left(-\dfrac{3\pi}{4}\right)\).

Exam style+20 XP

Find the exact value of \(\operatorname{cosec}^{2}\dfrac{\pi}{3} + \cot^{2}\dfrac{\pi}{3}\).

Skill unlocked

Six ratios, every angle, exactly.

This skill is now recorded on your profile. Keep the chain going.

-Rank -Level -Score -Topics
Go deeper · practice
Six original Cambridge-style questions
Reciprocal exact values in degrees and radians, ratios of angles in every quadrant including negative and large angles, finding all six ratios from one ratio and a quadrant, and a short exact-value proof.