← Trigonometric Functions (ME-T1, T2, T3)

NSWMaths Extension 1Syllabus dot point

How do we expand $\sin(A \pm B)$ , $\cos(A \pm B)$ and $\tan(A \pm B)$ , and what are these identities used for?

Use the sum and difference identities for sine, cosine and tangent to expand or simplify trigonometric expressions

A focused answer to the HSC Maths Extension 1 dot point on sum and difference identities. The expansions of $\sin(A \pm B)$ , $\cos(A \pm B)$ and $\tan(A \pm B)$ , derivation of double-angle and half-angle formulas, and exact values for non-standard angles, with worked examples.

Generated by Claude OpusReviewed by Better Tuition Academy8 min answerUpdated 2026-05-20

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What this dot point is asking

NESA wants you to know the sum and difference identities for sine, cosine and tangent, deploy them to expand or simplify expressions, and use them to compute exact values for non-standard angles like $15^\circ$ or $75^\circ$ .

The answer

The sum and difference identities

For sine,

DMATH_2

\sin(A - B) = \sin A \cos B - \cos A \sin B.

For cosine,

DMATH_4

\cos(A - B) = \cos A \cos B + \sin A \sin B.

Note the sign flip relative to the input.

For tangent,

DMATH_6

\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}.

The denominator must be non-zero; otherwise $\tan(A \pm B)$ is undefined.

Derivations of double-angle from sum identities

Setting $A = B = \theta$ :

\sin 2\theta = \sin(\theta + \theta) = 2 \sin \theta \cos \theta.

\cos 2\theta = \cos(\theta + \theta) = \cos^2 \theta - \sin^2 \theta.

\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}.

Half-angle identities

From the double-angle forms of $\cos 2\theta$ ,

\sin^2 \theta = \frac{1 - \cos 2\theta}{2}, \qquad \cos^2 \theta = \frac{1 + \cos 2\theta}{2}.

Taking square roots (and choosing the sign based on the quadrant of $\theta$ ):

\sin \frac{\phi}{2} = \pm \sqrt{\frac{1 - \cos \phi}{2}}, \qquad \cos \frac{\phi}{2} = \pm \sqrt{\frac{1 + \cos \phi}{2}}.

These are useful for finding exact values like $\cos 15^\circ$ .

Exact values for non-standard angles

The standard angles $0, 30, 45, 60, 90^\circ$ have well-known exact values. Combining them with sum and difference identities gives exact values for $15, 75, 105, 165^\circ$ and so on.

Standard exact values:

DMATH_13
DMATH_14

\sin 60^\circ = \frac{\sqrt{3}}{2}, \cos 60^\circ = \frac{1}{2}, \tan 60^\circ = \sqrt{3}.

Worked example

Compute an exact value using a difference

Find $\sin 15^\circ$ .

$15^\circ = 45^\circ - 30^\circ$ .

$\sin 15^\circ = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} - \sqrt{2}}{4}$ .

Use tan sum identity

Find $\tan 75^\circ$ .

$75^\circ = 45^\circ + 30^\circ$ .

$\tan 75^\circ = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ} = \frac{1 + 1/\sqrt{3}}{1 - 1/\sqrt{3}}$ .

Multiply top and bottom by $\sqrt{3}$ : $\frac{\sqrt{3} + 1}{\sqrt{3} - 1}$ .

Rationalise: $\frac{(\sqrt{3} + 1)^2}{(\sqrt{3})^2 - 1} = \frac{3 + 2\sqrt{3} + 1}{2} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$ .

Simplify a sum

Simplify $\sin(\theta + \frac{\pi}{4}) + \sin(\theta - \frac{\pi}{4})$ .

Expand each: $\sin \theta \cos \frac{\pi}{4} + \cos \theta \sin \frac{\pi}{4} + \sin \theta \cos \frac{\pi}{4} - \cos \theta \sin \frac{\pi}{4}$ .

The $\cos \theta$ terms cancel. Sum: $2 \sin \theta \cos \frac{\pi}{4} = \sqrt{2} \sin \theta$ .

Prove an identity

Prove $\cos(\theta + 60^\circ) + \cos(\theta - 60^\circ) = \cos \theta$ .

LHS: $\cos \theta \cos 60^\circ - \sin \theta \sin 60^\circ + \cos \theta \cos 60^\circ + \sin \theta \sin 60^\circ$ .

$\sin$ terms cancel. Sum: $2 \cos \theta \cos 60^\circ = 2 \cos \theta \cdot \frac{1}{2} = \cos \theta$ .

Combine with double-angle

Given $\sin \theta = \frac{3}{5}$ in Q1, find $\sin 3\theta$ .

$\sin 3\theta = \sin(2 \theta + \theta) = \sin 2\theta \cos \theta + \cos 2\theta \sin \theta$ .

$\sin 2\theta = 2 \cdot \frac{3}{5} \cdot \frac{4}{5} = \frac{24}{25}$ .

$\cos 2\theta = 1 - 2 \sin^2 \theta = 1 - \frac{18}{25} = \frac{7}{25}$ .

$\sin 3\theta = \frac{24}{25} \cdot \frac{4}{5} + \frac{7}{25} \cdot \frac{3}{5} = \frac{96}{125} + \frac{21}{125} = \frac{117}{125}$ .

Common traps

Sign mistake on $\cos(A + B)$: IMATH_1 has a minus, while $\sin(A + B) = \sin A \cos B + \cos A \sin B$ has a plus. The minus is unusual; mark it carefully.
Wrong quadrant for $\cos \beta$ or $\sin \beta$: When the question specifies a quadrant, get the sign of the other trig ratio right before substituting.
Forgetting to rationalise: A final answer of $\frac{1}{\sqrt{3} - 1}$ should be rationalised to $\frac{\sqrt{3} + 1}{2}$ . Markers expect simplified surds.
Tangent sum with zero denominator: IMATH_7 would have $1 - 1 \cdot 1 = 0$ in the denominator; the identity is undefined and you must reason via $\sin/\cos$ directly.
Reusing the wrong angle: IMATH_10 , not $\sin(75 + 30)^\circ$ in the most efficient form. Choose decompositions into standard angles.

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

2022 HSC Q93 marksUse a sum or difference identity to find the exact value of

\cos 75^\circ

Show worked answer →

Write $75^\circ = 45^\circ + 30^\circ$ and use $\cos(A + B) = \cos A \cos B - \sin A \sin B$ .

$\cos 75^\circ = \cos 45^\circ \cos 30^\circ - \sin 45^\circ \sin 30^\circ$ .

$= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2}$

$= \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}$ .

Markers reward identifying a useful decomposition into standard angles, the correct sum identity, and a final exact answer.

2020 HSC Q113 marksIf

\sin \alpha = \frac{3}{5}

with

\alpha

in the first quadrant and

\cos \beta = -\frac{12}{13}

with

\beta

in the second quadrant, find

\sin(\alpha + \beta)

Show worked answer →

In Q1: $\cos \alpha = \sqrt{1 - 9/25} = \frac{4}{5}$ .

In Q2: $\sin \beta = \sqrt{1 - 144/169} = \frac{5}{13}$ (positive in Q2).

Apply $\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$ .

$= \frac{3}{5} \cdot \left( -\frac{12}{13} \right) + \frac{4}{5} \cdot \frac{5}{13}$

$= -\frac{36}{65} + \frac{20}{65} = -\frac{16}{65}$ .

Markers reward correct quadrant signs for $\cos \alpha$ and $\sin \beta$ , the sum identity, and clean arithmetic.