Exact Trigonometric Values

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.

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° = √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 θ	cos θ	tan θ
0°	0	0	1	0
30°	π/6	1/2	√3/2	1/√3 = √3/3
45°	π/4	√2/2 = 1/√2	√2/2 = 1/√2	1
60°	π/3	√3/2	1/2	√3
90°	π/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) = √(2² − 1²) = √3

From the left-hand right-angled triangle:

For 30° (the angle at the top):

sin 30° = opposite/hypotenuse = 1/2
cos 30° = adjacent/hypotenuse = √3/2
tan 30° = opposite/adjacent = 1/√3 = √3/3

For 60° (the angle at the base):

sin 60° = opposite/hypotenuse = √3/2
cos 60° = adjacent/hypotenuse = 1/2
tan 60° = opposite/adjacent = √3/1 = √3

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 θ = cos(90° − θ).

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:

√(1² + 1²) = √2

Both acute angles are 45°.

For 45°:

sin 45° = opposite/hypotenuse = 1/√2 = √2/2
cos 45° = adjacent/hypotenuse = 1/√2 = √2/2
tan 45° = opposite/adjacent = 1/1 = 1

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 θ measured anticlockwise from the positive x-axis, the point on the unit circle is:

(cos θ, sin θ)

This means:

cos θ is the x-coordinate
sin θ is the y-coordinate

Values at 0°, 90°, 180°, 270°, 360°

From the unit circle:

θ	Point on circle	cos θ	sin θ	tan θ
0° (0)	(1, 0)	1	0	0
90° (π/2)	(0, 1)	0	1	undefined
180° (π)	(−1, 0)	−1	0	0
270° (3π/2)	(0, −1)	0	−1	undefined
360° (2π)	(1, 0)	1	0	0

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

Using the CAST Diagram

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

        S  |  A
     (sin+)| (all+)
    -------+-------
        T  |  C
     (tan+)| (cos+)

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° = sin(180° − 30°) = sin 30° = 1/2

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

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

cos 240° = −cos(240° − 180°) = −cos 60° = −1/2

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

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

tan 315° = −tan(360° − 315°) = −tan 45° = −1

Exact Values in Radians

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

sin(π/6) = 1/2
cos(π/3) = 1/2
tan(π/4) = 1
sin(π/3) = √3/2
cos(π/6) = √3/2
sin(π/4) = √2/2

Example 4: Find the exact value of cos(5π/6).

5π/6 is in the second quadrant (between π/2 and π), so cos is negative.

cos(5π/6) = −cos(π − 5π/6) = −cos(π/6) = −√3/2

Example 5: Find the exact value of sin(7π/6).

7π/6 is in the third quadrant (between π and 3π/2), so sin is negative.

sin(7π/6) = −sin(7π/6 − π) = −sin(π/6) = −1/2

Worked Problems

Problem 1: Without a calculator, evaluate 2sin(π/3) + cos(π/6).

= 2 × √3/2 + √3/2
= √3 + √3/2
= 2√3/2 + √3/2
= 3√3/2

Problem 2: Find the exact value of sin²(π/4) + cos²(π/3).

= (√2/2)² + (1/2)²
= 2/4 + 1/4
= 1/2 + 1/4
= 3/4

Problem 3: Show that tan 60° − tan 30° = 2tan 30°.

LHS = √3 − 1/√3 = √3 − √3/3 = 3√3/3 − √3/3 = 2√3/3
RHS = 2 × 1/√3 = 2/√3 = 2√3/3
LHS = RHS ∎

Common Errors to Avoid

Confusing sin 30° and sin 60°: Remember sin 30° = 1/2 (the smaller value for the smaller angle) and sin 60° = √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° = √2/2 (rationalised) rather than 1/√2, unless the question accepts either form.

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° = √0/2 = 0, sin 30° = √1/2 = 1/2, sin 45° = √2/2, sin 60° = √3/2, sin 90° = √4/2 = 1. This pattern (√0, √1, √2, √3, √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.

Total: 6 marks split AO1 = 5, AO2 = 1. This is an AO1-dominated question — AQA uses exact-value items chiefly to test procedural recall and unit-circle symmetry, with the AO2 mark reserved for the closing "no decimals" presentation discipline.

Synoptic links

Connects to:

Section F — Compound and double angle formulae: non-standard angles like $75°$ are evaluated by writing $75° = 45° + 30°$ and applying $\sin(A + B) = \sin A\cos B + \cos A\sin B$ with the standard exact values. The compound-angle formulae are only useful because the constituent values $\sin(45°), \cos(30°), \ldots$ are known exactly.
Section B — Surds and indices: every standard exact value lives in $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ . Manipulating expressions like $\sin(\pi/3)\cos(\pi/4) - \cos(\pi/3)\sin(\pi/4) = \dfrac{\sqrt{3}}{2} \cdot \dfrac{\sqrt{2}}{2} - \dfrac{1}{2} \cdot \dfrac{\sqrt{2}}{2} = \dfrac{\sqrt{6} - \sqrt{2}}{4}$ is pure surd arithmetic dressed up in trigonometric clothing.
Section E — Trigonometric identities: verifying identities like $\sin^2(\pi/6) + \cos^2(\pi/6) = 1$ provides instant sanity checks: $(1/2)^2 + (\sqrt{3}/2)^2 = 1/4 + 3/4 = 1$ . This Pythagorean check is the fastest way to spot a sign error mid-question.
Section H — Differentiation of trig functions: the standard derivatives $\frac{d}{dx}\sin x = \cos x$ require radian arguments. Substituting in $x = \pi/3$ gives a tangent-gradient of $\cos(\pi/3) = 1/2$ at that point — a calculation impossible without exact-value recall.

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

Worked Problems

Common Errors to Avoid

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

Synoptic links

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