Expand sin(x + 30^) using the addition formula. [2 marks]

Write sin x + sqrt(3)cos x in the form ksin(x + α). [3 marks]

SQA SQA Higher 2019-style practice: Given that sin A = 35 where A is acute, find the exact value of sin 2A and cos 2A.

Find cos A first. Since sin A = 35 and A is acute, cos A = sqrt(1 - sin^2 A) = 1 - 925 = 1625 = 45 (1 mark). sin 2A = 2sin Acos A = 2 · 35 · 45 = 2425 (1 mark). cos 2A = 1 - 2sin^2 A = 1 - 2(925) = 1 - 1825 = 725 (2 marks). Markers reward cos A from the identity, sin 2A from the double-angle formula, and cos 2A from any valid form.

SQA SQA Higher 2021-style practice: Express sqrt(3)sin x - cos x in the form ksin(x + α), where k > 0 and 0 α < 2π, and hence state the maximum value of the expression and the value of x at which it first occurs.

Amplitude: k = sqrt((sqrt(3))^2 + (-1)^2) = sqrt(3 + 1) = 2 (1 mark). Expand ksin(x + α) = ksin xcosα + kcos xsinα and compare: kcosα = sqrt(3) and ksinα = -1 (1 mark). So cosα = sqrt(3)2 > 0 and sinα = -12 < 0, placing α in the fourth quadrant: α = 2π - π6 = 11π6 (2 marks). So sqrt(3)sin x - cos x = 2sin\!(x + 11π6). The maximum value is k = 2, occurring when sin(x + α) = 1, i.e. x + 11π6 = π2, giving x = π2 - 11π6 = -8π6 = -4π3, or adding 2π, x = 2π3 (1 mark). Markers reward k, the quadrant for α from both signs, and the maximum with its location.

ScotlandMathsSyllabus dot point

How do the compound angle formulae let you expand and simplify trigonometric expressions, and how does the wave function combine a sine and a cosine into one?

The addition (compound angle) formulae for sine and cosine, the double angle formulae, their use in proving identities and solving equations, and the wave function that expresses a sin x plus b cos x in the form k sin of x plus a.

A focused answer to the SQA Higher Mathematics addition formulae content, covering the compound and double angle formulae for sine and cosine, their use in proving identities and solving equations, and the wave function that writes a sin x plus b cos x as a single sine wave.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

The compound angle formulae are $\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B$ and $\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B$ . The double angle formulae follow by setting $B = A$ : $\sin 2A = 2\sin A\cos A$ and $\cos 2A = \cos^2 A - \sin^2 A$ . The wave function writes $a\sin x + b\cos x$ as $k\sin(x + \alpha)$ , where $k = \sqrt{a^2 + b^2}$ and the auxiliary angle $\alpha$ satisfies $\tan\alpha = \dfrac{b}{a}$ , placed in the correct quadrant from the signs of $a$ and $b$ .

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What this dot point is asking
The compound and double angle formulae
The wave function
Examples in context
Try this

What this dot point is asking

The SQA wants you to use the compound angle (addition) formulae for sine and cosine, apply the double angle formulae, prove identities and solve equations with them, and express $a\sin x + b\cos x$ as a single wave in the form $k\sin(x + \alpha)$ .

The compound and double angle formulae

The compound angle formulae expand the sine or cosine of a sum or difference of two angles. Setting the two angles equal collapses them into the double angle formulae, which are the workhorse for proving identities, evaluating exact values and reducing equations.

The three forms of $\cos 2A$ let you choose the one that matches the rest of an equation: use $2\cos^2 A - 1$ when the equation involves $\cos A$ , and $1 - 2\sin^2 A$ when it involves $\sin A$ .

Find the exact value of

\cos 75^\circ

Find $\cos 75^\circ$ exactly, without a calculator, by splitting the angle into standard values.

Step 1: Write $75^\circ$ as a sum of standard angles

The exact values are known for $30^\circ$ , $45^\circ$ and $60^\circ$ . Since $75^\circ = 45^\circ + 30^\circ$ , the cosine addition formula applies:

\cos 75^\circ = \cos(45^\circ + 30^\circ)

Step 2: Apply the cosine addition formula

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

Note the minus sign: this distinguishes the cosine formula from the sine formula.

\cos(45^\circ + 30^\circ) = \cos 45^\circ\cos 30^\circ - \sin 45^\circ\sin 30^\circ

Step 3: Substitute the exact values and simplify

= \dfrac{1}{\sqrt{2}} \cdot \dfrac{\sqrt{3}}{2} - \dfrac{1}{\sqrt{2}} \cdot \dfrac{1}{2} = \dfrac{\sqrt{3}}{2\sqrt{2}} - \dfrac{1}{2\sqrt{2}} = \dfrac{\sqrt{3} - 1}{2\sqrt{2}}

Final answer: $\cos 75^\circ = \dfrac{\sqrt{3} - 1}{2\sqrt{2}}$ .

The wave function

The wave function rewrites a sum $a\sin x + b\cos x$ as a single sine wave $k\sin(x + \alpha)$ , which makes the amplitude and the location of the maximum obvious. You find $k$ from $\sqrt{a^2 + b^2}$ and the auxiliary angle $\alpha$ by matching the expanded form, taking care to place $\alpha$ in the quadrant fixed by the signs of $a$ and $b$ .

Express

3\sin x + 4\cos x

in the form

k\sin(x + \alpha)

Rewrite $3\sin x + 4\cos x$ as a single sine wave.

Step 1: Find the amplitude $k$

The amplitude $k$ is found from the Pythagorean combination of the two coefficients. This comes from expanding $k\sin(x + \alpha)$ and matching, but the result is always:

k = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5

Step 2: Match coefficients to find $\alpha$

Expanding $k\sin(x + \alpha) = k\sin x\cos\alpha + k\cos x\sin\alpha$ and comparing with $3\sin x + 4\cos x$ gives two equations:

k\cos\alpha = 3 \implies \cos\alpha = \dfrac{3}{5} \qquad k\sin\alpha = 4 \implies \sin\alpha = \dfrac{4}{5}

Step 3: Determine the quadrant and find $\alpha$

Both $\cos\alpha$ and $\sin\alpha$ are positive, so $\alpha$ lies in the first quadrant. Using $\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha} = \dfrac{4}{3}$ :

\alpha = \tan^{-1}\!\left(\dfrac{4}{3}\right) \approx 53.1^\circ

Final answer: $3\sin x + 4\cos x = 5\sin(x + 53.1^\circ)$ . The maximum value of the expression is $k = 5$ .

The wave form makes the maximum value ( $k$ ) and where it occurs easy to read off: the maximum is $k$ , reached when the bracket equals $90^\circ$ (or $\dfrac{\pi}{2}$ ).

Solve an equation using the double angle formula

Solve $\sin 2x = \sin x$ for $0 \le x < 360^\circ$ .

Step 1: Replace $\sin 2x$ using the double angle formula

The double angle formula $\sin 2x = 2\sin x\cos x$ converts the equation into one involving only $x$ :

2\sin x\cos x = \sin x

Step 2: Rearrange and factorise rather than divide

Bring all terms to one side: $2\sin x\cos x - \sin x = 0$ . Factor out $\sin x$ :

\sin x(2\cos x - 1) = 0

Dividing both sides by $\sin x$ would be wrong here: it silently discards any solution where $\sin x = 0$ . Factorising keeps all solutions visible.

Step 3: Solve each factor in the given interval

For $\sin x = 0$ : $x = 0^\circ$ or $x = 180^\circ$ .

For $2\cos x - 1 = 0$ , that is $\cos x = \dfrac{1}{2}$ : $x = 60^\circ$ or $x = 300^\circ$ .

Final answer: $x = 0^\circ, 60^\circ, 180^\circ, 300^\circ$ .

Examples in context

The wave function combines two oscillations of the same frequency into one. If two forces vary as $3\sin\omega t$ and $4\cos\omega t$ , their sum is $5\sin(\omega t + 53.1^\circ)$ , a single oscillation of amplitude $5$ . An engineer reads the resultant amplitude directly as $k = \sqrt{3^2 + 4^2} = 5$ and the phase lead as $\alpha$ , without resolving components every cycle.

Try this

Q1. Expand $\sin(x + 30^\circ)$ using the addition formula. [2 marks]

Cue. $\sin x\cos 30^\circ + \cos x\sin 30^\circ = \dfrac{\sqrt{3}}{2}\sin x + \dfrac{1}{2}\cos x$ .

Q2. Write $\sin x + \sqrt{3}\cos x$ in the form $k\sin(x + \alpha)$ . [3 marks]

Cue. $k = \sqrt{1 + 3} = 2$ , $\tan\alpha = \sqrt{3}$ , so $2\sin(x + 60^\circ)$ .

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA Higher 20194 marksGiven that

\sin A = \dfrac{3}{5}

where

A

is acute, find the exact value of

\sin 2A

and

\cos 2A

Show worked answer →

Find $\cos A$ first. Since $\sin A = \dfrac{3}{5}$ and $A$ is acute, $\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \dfrac{9}{25}} = \sqrt{\dfrac{16}{25}} = \dfrac{4}{5}$ (1 mark).

$\sin 2A = 2\sin A\cos A = 2 \cdot \dfrac{3}{5} \cdot \dfrac{4}{5} = \dfrac{24}{25}$ (1 mark).

$\cos 2A = 1 - 2\sin^2 A = 1 - 2\left(\dfrac{9}{25}\right) = 1 - \dfrac{18}{25} = \dfrac{7}{25}$ (2 marks). Markers reward $\cos A$ from the identity, $\sin 2A$ from the double-angle formula, and $\cos 2A$ from any valid form.

SQA Higher 20215 marksExpress

\sqrt{3}\sin x - \cos x

in the form

k\sin(x + \alpha)

, where

k > 0

and

0 \le \alpha < 2\pi

, and hence state the maximum value of the expression and the value of

x

at which it first occurs.

Show worked answer →

Amplitude: $k = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = 2$ (1 mark).

Expand $k\sin(x + \alpha) = k\sin x\cos\alpha + k\cos x\sin\alpha$ and compare: $k\cos\alpha = \sqrt{3}$ and $k\sin\alpha = -1$ (1 mark). So $\cos\alpha = \dfrac{\sqrt{3}}{2} > 0$ and $\sin\alpha = -\dfrac{1}{2} < 0$ , placing $\alpha$ in the fourth quadrant: $\alpha = 2\pi - \dfrac{\pi}{6} = \dfrac{11\pi}{6}$ (2 marks).

So $\sqrt{3}\sin x - \cos x = 2\sin\!\left(x + \dfrac{11\pi}{6}\right)$ . The maximum value is $k = 2$ , occurring when $\sin(x + \alpha) = 1$ , i.e. $x + \dfrac{11\pi}{6} = \dfrac{\pi}{2}$ , giving $x = \dfrac{\pi}{2} - \dfrac{11\pi}{6} = -\dfrac{8\pi}{6} = -\dfrac{4\pi}{3}$ , or adding $2\pi$ , $x = \dfrac{2\pi}{3}$ (1 mark). Markers reward $k$ , the quadrant for $\alpha$ from both signs, and the maximum with its location.

Related dot points

Sources & how we know this

SQA Higher Mathematics Course Specification — SQA (2018)