Exact Trigonometric Values

Builds on: aqa-alevel-maths-pure-1 / trigonometry — this lesson is the A2-level deep dive.

This lesson covers the exact values of $\sin$ , $\cos$ , and $\tan$ for the standard angles $0°$ , $30°$ , $45°$ , $60°$ , and $90°$ (and their radian equivalents). These values appear throughout A-Level Mathematics — in trigonometric equations, calculus, coordinate geometry, and mechanics. You are expected to recall them without a calculator.

Spec Mapping — AQA 7357 Section E Trigonometry. This lesson covers the exact surd values of sine, cosine and tangent at special angles content of Section E, at the depth required for A2-level synoptic questions. Refer to the official AQA specification document for exact wording.

Why Exact Values Matter

In many exam questions — especially "show that" and "prove" questions — you must use exact values rather than decimal approximations. Writing $\sin 60° = 0.866$ will not earn full marks; you need $\sin 60° = \tfrac{\sqrt{3}}{2}$ .

Exact values also arise naturally when solving trigonometric equations, and you will use them extensively in the addition and double angle formulae covered in later lessons.

The Standard Exact Values

Table of Exact Values

Angle (°)	Angle (rad)	$\sin\theta$	$\cos\theta$	$\tan\theta$
$0°$	$0$	$0$	$1$	$0$
$30°$	$\tfrac{\pi}{6}$	$\tfrac{1}{2}$	$\tfrac{\sqrt{3}}{2}$	$\tfrac{1}{\sqrt{3}} = \tfrac{\sqrt{3}}{3}$
$45°$	$\tfrac{\pi}{4}$	$\tfrac{\sqrt{2}}{2} = \tfrac{1}{\sqrt{2}}$	$\tfrac{\sqrt{2}}{2} = \tfrac{1}{\sqrt{2}}$	$1$
$60°$	$\tfrac{\pi}{3}$	$\tfrac{\sqrt{3}}{2}$	$\tfrac{1}{2}$	$\sqrt{3}$
$90°$	$\tfrac{\pi}{2}$	$1$	$0$	undefined

You must memorise this table. Every entry is examinable.

Deriving from the Equilateral Triangle (30° and 60°)

Consider an equilateral triangle with side length 2. All angles are 60°. Drop a perpendicular from the top vertex to the base. This creates two right-angled triangles with:

Hypotenuse = 2
Base (short side) = 1
Height (long side) = $\sqrt{2^2 - 1^2} = \sqrt{3}$

From the left-hand right-angled triangle:

For 30° (the angle at the top):

\begin{aligned} \sin 30^{\circ} &= \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{2} \\ \cos 30^{\circ} &= \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\sqrt{3}}{2} \\ \tan 30^{\circ} &= \frac{\text{opposite}}{\text{adjacent}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \end{aligned}

For 60° (the angle at the base):

\begin{aligned} \sin 60^{\circ} &= \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\sqrt{3}}{2} \\ \cos 60^{\circ} &= \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{2} \\ \tan 60^{\circ} &= \frac{\text{opposite}}{\text{adjacent}} = \sqrt{3} \end{aligned}

Notice that $\sin 30° = \cos 60°$ and $\cos 30° = \sin 60°$ . This is because $30°$ and $60°$ are complementary angles (they sum to $90°$ ), and in general $\sin\theta = \cos(90° - \theta)$ .

Deriving from the Isosceles Right-Angled Triangle (45°)

Consider a right-angled isosceles triangle with the two equal sides of length 1. The hypotenuse has length:

\sqrt{1^{2} + 1^{2}} = \sqrt{2}

Both acute angles are 45°.

For 45°:

\begin{aligned} \sin 45^{\circ} &= \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \\ \cos 45^{\circ} &= \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \\ \tan 45^{\circ} &= \frac{\text{opposite}}{\text{adjacent}} = 1 \end{aligned}

Notice $\sin 45° = \cos 45°$ , which makes sense because both acute angles in an isosceles right-angled triangle are equal.

The Unit Circle Approach

The unit circle has centre $(0, 0)$ and radius 1. For any angle $\theta$ measured anticlockwise from the positive $x$ -axis, the point on the unit circle is:

$(\cos\theta,\ \sin\theta)$

This means:

$\cos\theta$ is the $x$ -coordinate
$\sin\theta$ is the $y$ -coordinate

Values at 0°, 90°, 180°, 270°, 360°

From the unit circle:

$\theta$	Point on circle	$\cos\theta$	$\sin\theta$	$\tan\theta$
$0°$ ( $0$ )	$(1, 0)$	$1$	$0$	$0$
$90°$ ( $\tfrac{\pi}{2}$ )	$(0, 1)$	$0$	$1$	undefined
$180°$ ( $\pi$ )	$(-1, 0)$	$-1$	$0$	$0$
$270°$ ( $\tfrac{3\pi}{2}$ )	$(0, -1)$	$0$	$-1$	undefined
$360°$ ( $2\pi$ )	$(1, 0)$	$1$	$0$	$0$

$\tan\theta = \sin\theta / \cos\theta$ , so $\tan\theta$ is undefined when $\cos\theta = 0$ (at $90°$ and $270°$ ).

Using the CAST Diagram

The CAST diagram tells you which trigonometric functions are positive in each quadrant:

Reading anticlockwise from the fourth quadrant: Cos, All, Sin, Tan — or the mnemonic "Cast" starting bottom-right.

First quadrant ( $0°$ to $90°$ ): All positive
Second quadrant ( $90°$ to $180°$ ): Only $\sin$ positive
Third quadrant ( $180°$ to $270°$ ): Only $\tan$ positive
Fourth quadrant ( $270°$ to $360°$ ): Only $\cos$ positive

Example 1: Find the exact value of $\sin 150°$ .

$150°$ is in the second quadrant, so $\sin$ is positive.

\sin 150^{\circ} = \sin(180^{\circ} - 30^{\circ}) = \sin 30^{\circ} = 1/2

Example 2: Find the exact value of $\cos 240°$ .

$240°$ is in the third quadrant, so $\cos$ is negative.

\cos 240^{\circ} = -\cos(240^{\circ} - 180^{\circ}) = -\cos 60^{\circ} = -1/2

Example 3: Find the exact value of $\tan 315°$ .

$315°$ is in the fourth quadrant, so $\tan$ is negative.

\tan 315^{\circ} = -\tan(360^{\circ} - 315^{\circ}) = -\tan 45^{\circ} = -1

Exact Values in Radians

Since A-Level questions often use radians, you need to be equally comfortable with:

\begin{aligned} \sin( \pi /6) = 1/2 \\ \cos( \pi /3) = 1/2 \\ \tan( \pi /4) = 1 \\ \sin( \pi /3) = \sqrt{3}/2 \\ \cos( \pi /6) = \sqrt{3}/2 \\ \sin( \pi /4) = \sqrt{2}/2 \end{aligned}

Example 4: Find the exact value of $\cos(\tfrac{5\pi}{6})$ .

$\tfrac{5\pi}{6}$ is in the second quadrant (between $\tfrac{\pi}{2}$ and $\pi$ ), so $\cos$ is negative.

\cos(5 \pi /6) = -\cos( \pi - 5 \pi /6) = -\cos( \pi /6) = - \sqrt{3}/2

Example 5: Find the exact value of $\sin(\tfrac{7\pi}{6})$ .

$\tfrac{7\pi}{6}$ is in the third quadrant (between $\pi$ and $\tfrac{3\pi}{2}$ ), so $\sin$ is negative.

\sin(7 \pi /6) = -\sin(7 \pi /6 - \pi ) = -\sin( \pi /6) = -1/2

Specimen Question (Modelled on the AQA Paper Format)

Note: this question is constructed to model AQA Paper 1/2/3 style; it is not a reproduction of any published past paper.

Worked Problems

Problem 1: Without a calculator, evaluate $2\sin(\tfrac{\pi}{3}) + \cos(\tfrac{\pi}{6})$ .

\begin{aligned} &= 2 \times \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \\ &= \sqrt{3} + \frac{\sqrt{3}}{2} \\ &= \frac{2\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \\ &= \frac{3\sqrt{3}}{2} \end{aligned}

Problem 2: Find the exact value of $\sin^2(\tfrac{\pi}{4}) + \cos^2(\tfrac{\pi}{3})$ .

\begin{aligned} &= \left(\frac{\sqrt{2}}{2}\right)^{2} + \left(\frac{1}{2}\right)^{2} \\ &= \frac{2}{4} + \frac{1}{4} \\ &= \frac{1}{2} + \frac{1}{4} \\ &= \frac{3}{4} \end{aligned}

Problem 3: Show that $\tan 60° - \tan 30° = 2\tan 30°$ .

\begin{aligned} \text{LHS} &= \sqrt{3} - \frac{1}{\sqrt{3}} = \sqrt{3} - \frac{\sqrt{3}}{3} = \frac{3\sqrt{3}}{3} - \frac{\sqrt{3}}{3} = \frac{2\sqrt{3}}{3} \\ \text{RHS} &= 2 \times \frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \\ \text{LHS} &= \text{RHS} \quad \blacksquare \end{aligned}

Common Errors to Avoid

Confusing $\sin 30°$ and $\sin 60°$ : Remember $\sin 30° = \tfrac{1}{2}$ (the smaller value for the smaller angle) and $\sin 60° = \tfrac{\sqrt{3}}{2}$ .
Forgetting the negative signs in other quadrants: Always use the CAST diagram.
Writing $\tan 90° = 1$ : $\tan 90°$ is undefined, not 1.
Rationalising: When asked for exact values, give $\sin 45° = \tfrac{\sqrt{2}}{2}$ (rationalised) rather than $\tfrac{1}{\sqrt{2}}$ , unless the question accepts either form.

Synoptic Links

This topic connects to:

Solving trig equations (aqa-alevel-maths-trigonometry-depth / solving-trigonometric-equations) — exact values appear as principal solutions on non-calculator papers.
Addition formulae (aqa-alevel-maths-trigonometry-depth / addition-formulae) — exact values for $15°$ and $75°$ derive from $\sin(45° - 30°)$ and $\cos(45° + 30°)$ .
Integration of trig (aqa-alevel-maths-calculus-applications / definite-integration) — exact bounds in definite integrals frequently use $\tfrac{\pi}{6}$ , $\tfrac{\pi}{4}$ , $\tfrac{\pi}{3}$ .

Summary

Memorise the exact values of $\sin$ , $\cos$ , and $\tan$ for $0°$ , $30°$ , $45°$ , $60°$ , and $90°$ .
Derive 30° and 60° values from an equilateral triangle of side 2.
Derive 45° values from an isosceles right-angled triangle with equal sides 1.
Use the unit circle to understand values at 0°, 90°, 180°, 270°, and 360°.
Use the CAST diagram to determine signs in all four quadrants.
Be fluent in both degree and radian forms.

Exam Tip: These exact values are not in the formula booklet — you must learn them. A quick way to recall $\sin$ values: $\sin 0° = \tfrac{\sqrt{0}}{2} = 0$ , $\sin 30° = \tfrac{\sqrt{1}}{2} = \tfrac{1}{2}$ , $\sin 45° = \tfrac{\sqrt{2}}{2}$ , $\sin 60° = \tfrac{\sqrt{3}}{2}$ , $\sin 90° = \tfrac{\sqrt{4}}{2} = 1$ . This pattern ( $\sqrt{0}, \sqrt{1}, \sqrt{2}, \sqrt{3}, \sqrt{4}$ all divided by 2) is a useful memory aid. The $\cos$ values are the same sequence in reverse.

A-Level Deep Dive: Exact Trigonometric Values

Spec mapping

AQA 7357 specification, Paper 1 — Pure Mathematics, Section E (Trigonometry) covers exact values of $\sin$ , $\cos$ and $\tan$ for $0$ , $\pi/6$ , $\pi/4$ , $\pi/3$ , $\pi/2$ and multiples thereof; know and use exact values of $\sin$ and $\cos$ for these and other angles obtained via symmetry of the unit circle (refer to the official specification document for exact wording). Although this sub-strand sits in Section E, exact-value fluency feeds Section F (Compound and double angle formulae, e.g. evaluating $\sin(75°)$ ), Section H (Differentiation of trig functions, where $\frac{d}{dx}\sin x = \cos x$ requires radian arguments), and Section I (Integration with trig limits, e.g. $\int_0^{\pi/3} \cos x\,dx = \sin(\pi/3) - \sin 0 = \sqrt{3}/2$ ). The AQA formula booklet does not list these standard exact values — they must be memorised.

Worked example with full mark scheme

Question (8 marks):

(a) Without using a calculator, find the exact value of $\sin\!\left(\dfrac{2\pi}{3}\right) + \cos\!\left(\dfrac{7\pi}{6}\right)$ , giving your answer in the form $p\sqrt{3}$ where $p$ is a rational number to be found. (5)

(b) Hence, or otherwise, solve $2\sin\theta + 1 = 0$ for $\theta \in [0, 2\pi)$ , giving exact answers in radians. (3)

Solution with mark scheme:

(a) Step 1 — locate $2\pi/3$ on the unit circle.

$2\pi/3 = 120°$ lies in the second quadrant. The reference angle (acute angle to the $x$ -axis) is $\pi - 2\pi/3 = \pi/3$ . In the second quadrant, sine is positive and cosine is negative.

$\sin\!\left(\dfrac{2\pi}{3}\right) = +\sin\!\left(\dfrac{\pi}{3}\right) = \dfrac{\sqrt{3}}{2}$

M1 — identifying the correct quadrant and reference angle. Common error: students reach for a calculator in radian mode and read off a decimal — that earns nothing on a "without a calculator" instruction. The mark is for the symmetry argument, not the value alone.

A1 — correct exact value $\sqrt{3}/2$ , with the positive sign justified by quadrant.

Step 2 — locate $7\pi/6$ on the unit circle.

$7\pi/6 = 210°$ lies in the third quadrant. The reference angle is $7\pi/6 - \pi = \pi/6$ . In the third quadrant, both sine and cosine are negative.

$\cos\!\left(\dfrac{7\pi}{6}\right) = -\cos\!\left(\dfrac{\pi}{6}\right) = -\dfrac{\sqrt{3}}{2}$

M1 — third-quadrant identification with reference angle $\pi/6$ .

A1 — correct signed value $-\sqrt{3}/2$ .

Step 3 — combine.

$\sin\!\left(\dfrac{2\pi}{3}\right) + \cos\!\left(\dfrac{7\pi}{6}\right) = \dfrac{\sqrt{3}}{2} + \left(-\dfrac{\sqrt{3}}{2}\right) = 0$

But $0$ can be written as $0\sqrt{3}$ , so $p = 0$ .

A1 — final form $p = 0$ , presented in the requested form.

(b) Step 1 — rearrange.

$2\sin\theta + 1 = 0 \implies \sin\theta = -\tfrac{1}{2}$ .

M1 — isolating $\sin\theta$ .

Step 2 — use exact values.

$\sin(\pi/6) = 1/2$ , so $\sin\theta = -1/2$ requires the third or fourth quadrant. The reference angle is $\pi/6$ .

Third quadrant: $\theta = \pi + \pi/6 = 7\pi/6$ .
Fourth quadrant: $\theta = 2\pi - \pi/6 = 11\pi/6$ .

M1 — applying quadrant rules to a negative sine.

A1 — $\theta = 7\pi/6, 11\pi/6$ , both in radians within $[0, 2\pi)$ .

Total: 8 marks (M4 A4, split as shown).

Specimen question modelled on the AQA 7357 Paper 1 format

Question (6 marks): Given that $\theta$ is acute and $\cos\theta = \dfrac{1}{\sqrt{2}}$ :

(a) State the exact value of $\theta$ in radians. (1)

(b) Hence, find the exact value of $\sin(2\theta) + \tan(3\theta)$ , simplifying your answer. (5)

Mark scheme decomposition by AO:

(a)

B1 (AO1.1b) — $\theta = \pi/4$ (acute, so unique).

(b)

M1 (AO1.1a) — $2\theta = \pi/2$ , so $\sin(2\theta) = \sin(\pi/2) = 1$ .
M1 (AO1.1a) — $3\theta = 3\pi/4$ , in the second quadrant, with reference angle $\pi/4$ .
M1 (AO1.1b) — $\tan(3\pi/4) = -\tan(\pi/4) = -1$ (tangent negative in Q2).
A1 (AO1.1b) — $\sin(2\theta) + \tan(3\theta) = 1 + (-1) = 0$ .
A1 (AO2.5) — exact form, no decimal substitution at any stage.

Exact Trigonometric Values

Exact Trigonometric Values

Why Exact Values Matter

The Standard Exact Values

Table of Exact Values

Deriving from the Equilateral Triangle (30° and 60°)

Deriving from the Isosceles Right-Angled Triangle (45°)

The Unit Circle Approach

Values at 0°, 90°, 180°, 270°, 360°

Using the CAST Diagram

Exact Values in Radians

Specimen Question (Modelled on the AQA Paper Format)

Worked Problems

Common Errors to Avoid

Synoptic Links

Summary

A-Level Deep Dive: Exact Trigonometric Values

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the AQA 7357 Paper 1 format

More in Mathematics