Convert 135^ to radians. [1 mark]

A sector has radius 6\,cm and angle 0.5 radians. Find its area. [2 marks]

Find R if 5sinθ + 12cosθ = Rsin(θ + α). [2 marks]

WJEC WJEC A2 style-style practice: Express 3sinθ + 4cosθ in the form Rsin(θ + α) with R > 0 and 0 < α < 90^, then state the maximum value.

Match coefficients with the expansion Rsin(θ + α) = Rcosα sinθ + Rsinα cosθ. So Rcosα = 3 and Rsinα = 4. R = sqrt(3^2 + 4^2) = sqrt(25) = 5. tanα = 43, so α = 53.1^. Therefore 3sinθ + 4cosθ = 5sin(θ + 53.1^). The maximum value is R = 5, reached when sin(θ + α) = 1. Markers reward finding R by Pythagoras, α from the tangent ratio, the correct form, and the maximum value 5. Mixing up which coefficient gives cosα and which gives sinα is the usual error.

WJEC WJEC A2 style-style practice: Solve cos 2θ + sinθ = 0 for 0 θ < 2π, giving answers in radians.

Use the double-angle identity to write everything in terms of sinθ. cos 2θ = 1 - 2sin^2θ, so 1 - 2sin^2θ + sinθ = 0. Rearrange: 2sin^2θ - sinθ - 1 = 0, which factorises as (2sinθ + 1)(sinθ - 1) = 0. So sinθ = -12 or sinθ = 1. sinθ = 1 gives θ = π2. sinθ = -12 gives θ = 7π6 and θ = 11π6. Solutions: θ = π2, 7π6, 11π6. Markers reward the double-angle substitution, factorising the quadratic, and all solutions in radians within range.

WalesMathsSyllabus dot point

How do we use radians, the reciprocal ratios, and the compound and double-angle identities to simplify and solve trig problems?

Radian measure, the reciprocal functions secant, cosecant and cotangent, the compound and double-angle formulae, and the harmonic form R sin(theta + alpha).

A focused answer to WJEC A2 Unit 3 trigonometry, covering radian measure, the reciprocal functions secant, cosecant and cotangent, the compound and double-angle identities, and the harmonic form R sin(theta plus alpha).

Generated by Claude Opus 4.814 min answerUpdated 2026-06-16

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

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Quick answer

Radians: $\pi$ radians $= 180^{\circ}$ ; arc length $s = r\theta$ and sector area $= \tfrac{1}{2}r^2\theta$ . The reciprocal ratios are $\sec\theta = \dfrac{1}{\cos\theta}$ , $\csc\theta = \dfrac{1}{\sin\theta}$ , $\cot\theta = \dfrac{1}{\tan\theta}$ , with identities $1 + \tan^2\theta = \sec^2\theta$ and $1 + \cot^2\theta = \csc^2\theta$ . Compound angles: $\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B$ . Double angles: $\sin 2\theta = 2\sin\theta\cos\theta$ , $\cos 2\theta = 1 - 2\sin^2\theta$ . The harmonic form writes $a\sin\theta + b\cos\theta$ as $R\sin(\theta + \alpha)$ with $R = \sqrt{a^2 + b^2}$ .

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What this dot point is asking

WJEC wants you to use radian measure (including arc length and sector area), the reciprocal functions $\sec\theta$ , $\csc\theta$ and $\cot\theta$ with their identities, the compound-angle and double-angle formulae, and the harmonic form $R\sin(\theta + \alpha)$ for expressions like $a\sin\theta + b\cos\theta$ . These identities are essential for the calculus and differential equations later in the course.

The answer

Radian measure

A radian is the angle subtended at the centre by an arc equal in length to the radius. Since a full circle is $2\pi$ radians, $\pi$ radians $= 180^{\circ}$ .

Calculus on trig functions requires radians, so convert degrees with $\theta_{\text{rad}} = \theta_{\text{deg}} \times \dfrac{\pi}{180}$ .

Reciprocal functions

Compound and double-angle formulae

The three forms of $\cos 2\theta$ let you choose the one that simplifies an equation to a quadratic in a single ratio. The $\cos 2\theta = 2\cos^2\theta - 1$ form is also the key to integrating $\cos^2\theta$ .

The harmonic form R sin(theta + alpha)

Expressing $a\sin\theta + b\cos\theta$ as a single sinusoid makes the maximum, minimum and solutions immediate.

Write in harmonic form and solve

Express $\sin\theta - \sqrt{3}\cos\theta$ as $R\sin(\theta - \alpha)$ and hence solve $\sin\theta - \sqrt{3}\cos\theta = 1$ for $0 \le \theta < 360^{\circ}$ .

step Match coefficients

$R\sin(\theta - \alpha) = R\cos\alpha\sin\theta - R\sin\alpha\cos\theta$ , so $R\cos\alpha = 1$ and $R\sin\alpha = \sqrt{3}$ .

step Find R and alpha

$R = \sqrt{1 + 3} = 2$ and $\tan\alpha = \sqrt{3}$ , so $\alpha = 60^{\circ}$ . Thus the expression is $2\sin(\theta - 60^{\circ})$ .

step Solve the equation

$2\sin(\theta - 60^{\circ}) = 1$ , so $\sin(\theta - 60^{\circ}) = \tfrac{1}{2}$ , giving $\theta - 60^{\circ} = 30^{\circ}$ or $150^{\circ}$ .

step Find theta in range

$\theta = 90^{\circ}$ or $\theta = 210^{\circ}$ .

Examples in context

Example 1. Maximum of a model. A tide height is modelled by $h = 3\sin t + 4\cos t$ metres. Writing it as $5\sin(t + \alpha)$ shows the maximum height is $5\,\text{m}$ and the minimum is $-5\,\text{m}$ , reached when the sine equals $\pm 1$ . The harmonic form reads off the extremes instantly.

Example 2. Proving an identity. Show $\dfrac{\sin 2\theta}{1 + \cos 2\theta} = \tan\theta$ . The numerator is $2\sin\theta\cos\theta$ and the denominator is $1 + (2\cos^2\theta - 1) = 2\cos^2\theta$ , so the fraction is $\dfrac{2\sin\theta\cos\theta}{2\cos^2\theta} = \dfrac{\sin\theta}{\cos\theta} = \tan\theta$ . The double-angle identities collapse the expression cleanly.

Try this

Q1. Convert $135^{\circ}$ to radians. [1 mark]

Cue. $135 \times \dfrac{\pi}{180} = \dfrac{3\pi}{4}$ .

Q2. A sector has radius $6\,\text{cm}$ and angle $0.5$ radians. Find its area. [2 marks]

Cue. $A = \tfrac{1}{2}r^2\theta = \tfrac{1}{2}(36)(0.5) = 9\,\text{cm}^2$ .

Q3. Find $R$ if $5\sin\theta + 12\cos\theta = R\sin(\theta + \alpha)$ . [2 marks]

Cue. $R = \sqrt{5^2 + 12^2} = \sqrt{169} = 13$ .

Exam-style practice questions

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

WJEC A2 style6 marksExpress

3\sin\theta + 4\cos\theta

in the form

R\sin(\theta + \alpha)

with

R > 0

and

0 < \alpha < 90^{\circ}

, then state the maximum value.

Show worked answer →

Match coefficients with the expansion $R\sin(\theta + \alpha) = R\cos\alpha \sin\theta + R\sin\alpha \cos\theta$ .

So $R\cos\alpha = 3$ and $R\sin\alpha = 4$ .

$R = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$ .

$\tan\alpha = \dfrac{4}{3}$ , so $\alpha = 53.1^{\circ}$ .

Therefore $3\sin\theta + 4\cos\theta = 5\sin(\theta + 53.1^{\circ})$ .

The maximum value is $R = 5$ , reached when $\sin(\theta + \alpha) = 1$ .

Markers reward finding $R$ by Pythagoras, $\alpha$ from the tangent ratio, the correct form, and the maximum value $5$ . Mixing up which coefficient gives $\cos\alpha$ and which gives $\sin\alpha$ is the usual error.

WJEC A2 style5 marksSolve

\cos 2\theta + \sin\theta = 0

for

0 \le \theta < 2\pi

, giving answers in radians.

Show worked answer →

Use the double-angle identity to write everything in terms of $\sin\theta$ .

$\cos 2\theta = 1 - 2\sin^2\theta$ , so $1 - 2\sin^2\theta + \sin\theta = 0$ .

Rearrange: $2\sin^2\theta - \sin\theta - 1 = 0$ , which factorises as $(2\sin\theta + 1)(\sin\theta - 1) = 0$ .

So $\sin\theta = -\dfrac{1}{2}$ or $\sin\theta = 1$ .

$\sin\theta = 1$ gives $\theta = \dfrac{\pi}{2}$ .
$\sin\theta = -\dfrac{1}{2}$ gives $\theta = \dfrac{7\pi}{6}$ and $\theta = \dfrac{11\pi}{6}$ .

Solutions: $\theta = \dfrac{\pi}{2}, \dfrac{7\pi}{6}, \dfrac{11\pi}{6}$ . Markers reward the double-angle substitution, factorising the quadratic, and all solutions in radians within range.

Related dot points

Sources & how we know this

WJEC GCE AS/A Level Mathematics specification (from 2017) — WJEC (2017)