State the period of y = tanθ. [1 mark]

Solve tanθ = 1 for 0^ θ 360^. [2 marks]

CCEA CCEA 2020-style practice: Solve 2sin^2θ + sinθ - 1 = 0 for 0^ θ 360^.

Treat sinθ as the variable and factorise the quadratic: 2sin^2θ + sinθ - 1 = (2sinθ - 1)(sinθ + 1) = 0. So sinθ = (1)/(2) or sinθ = -1. For sinθ = (1)/(2): θ = 30^ and θ = 150^ (second-quadrant partner). For sinθ = -1: θ = 270^. So θ = 30^, 150^, 270^. Markers reward the factorisation, both values of sinθ, all the solutions in range, and not missing the 150^ partner.

CCEA CCEA 2018-style practice: In triangle ABC, AB = 7\,cm, AC = 9\,cm and angle BAC = 50^. Find the length BC and the area of the triangle.

Use the cosine rule with BC = a opposite the known angle: a^2 = 7^2 + 9^2 - 2(7)(9)cos 50^ = 49 + 81 - 126cos 50^. cos 50^ ≈ 0.6428, so a^2 ≈ 130 - 80.99 = 49.01, giving BC ≈ 7.00\,cm. Area = (1)/(2)(7)(9)sin 50^ = 31.5 × 0.766 ≈ 24.1\,cm^2. Markers reward the correct cosine-rule setup, the value of BC, the area formula (1)/(2)absin C, and the numerical area.

Northern IrelandMathsSyllabus dot point

How do trigonometric ratios, identities and graphs let us model and solve angle problems?

The trigonometric ratios and their graphs, the sine and cosine rules and the area of a triangle, the identities $\sin^2\theta + \cos^2\theta = 1$ and $\tan\theta = \sin\theta/\cos\theta$ , and solving trigonometric equations.

A CCEA A-Level Mathematics answer on the trigonometric ratios and their graphs, the sine and cosine rules and the area of a triangle, the Pythagorean identity, and solving trigonometric equations over a given interval.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-16

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

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What this dot point is asking

CCEA wants you to know the trigonometric ratios and the shapes of their graphs, apply the sine and cosine rules and the area formula to any triangle, use the Pythagorean identity and the $\tan = \sin/\cos$ identity to simplify and prove, and solve trigonometric equations (including quadratics in a ratio) over a stated interval. Trigonometry threads through coordinate geometry, calculus and mechanics.

The answer

The ratios and their graphs

The sine and cosine rules

Use the sine rule when you have a side and its opposite angle; use the cosine rule when you have two sides and the included angle (to find the third side) or all three sides (to find an angle). Watch for the ambiguous case of the sine rule, where two triangles may satisfy the data.

Identities and equations

To solve $\sin\theta = k$ , find the principal value with the inverse function, then use the symmetry of the graph to find every solution in the interval. Always work in the interval the question gives, and remember each value can have more than one solution.

Worked example: an equation using the identity

Solve a trig equation by converting to one ratio

Solve $2\cos^2\theta = 1 + \sin\theta$ for $0^\circ \le \theta \le 360^\circ$ .

step Replace cos squared

Using $\cos^2\theta = 1 - \sin^2\theta$ :

2(1 - \sin^2\theta) = 1 + \sin\theta.

step Form a quadratic in sin

2 - 2\sin^2\theta = 1 + \sin\theta \Rightarrow 2\sin^2\theta + \sin\theta - 1 = 0.

step Factorise and solve

(2\sin\theta - 1)(\sin\theta + 1) = 0 \Rightarrow \sin\theta = \tfrac{1}{2} \text{ or } \sin\theta = -1.

\theta = 30^\circ, 150^\circ

(from

\sin\theta = \tfrac{1}{2}

) and

\theta = 270^\circ

(from

\sin\theta = -1

Examples in context

Example 1. A surveying triangle. To find an inaccessible distance, surveyors measure a baseline and two angles, then apply the sine rule. Knowing one side and all the angles fixes the triangle, which is why the sine rule is the natural first tool.

Example 2. Modelling daylight hours. The number of daylight hours through the year is roughly $D = 12 + 4\sin\!\left(\frac{360t}{365}\right)$ , a sine wave of amplitude $4$ about a mean of $12$ . Reading amplitude and period from the graph is the same skill as sketching $y = \sin\theta$ .

Try this

Q1. State the period of $y = \tan\theta$ . [1 mark]

Cue. $180^\circ$ .

Q2. A triangle has sides $5$ , $6$ and $7$ . Find its largest angle. [3 marks]

Cue. Cosine rule on the side $7$ : $49 = 25 + 36 - 60\cos C$ , so $\cos C = 0.2$ and $C \approx 78.5^\circ$ .

Q3. Solve $\tan\theta = 1$ for $0^\circ \le \theta \le 360^\circ$ . [2 marks]

Cue. $\theta = 45^\circ$ and $\theta = 225^\circ$ .

Exam-style practice questions

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

CCEA 20206 marksSolve

2\sin^2\theta + \sin\theta - 1 = 0

for

0^\circ \le \theta \le 360^\circ

Show worked answer →

Treat $\sin\theta$ as the variable and factorise the quadratic:

$2\sin^2\theta + \sin\theta - 1 = (2\sin\theta - 1)(\sin\theta + 1) = 0.$

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

For $\sin\theta = \frac{1}{2}$ : $\theta = 30^\circ$ and $\theta = 150^\circ$ (second-quadrant partner).

For $\sin\theta = -1$ : $\theta = 270^\circ$ .

So $\theta = 30^\circ, 150^\circ, 270^\circ$ .

Markers reward the factorisation, both values of $\sin\theta$ , all the solutions in range, and not missing the $150^\circ$ partner.

CCEA 20185 marksIn triangle

ABC

AB = 7\,\text{cm}

AC = 9\,\text{cm}

and angle

BAC = 50^\circ

. Find the length

BC

and the area of the triangle.

Show worked answer →

Use the cosine rule with $BC = a$ opposite the known angle:

$a^2 = 7^2 + 9^2 - 2(7)(9)\cos 50^\circ = 49 + 81 - 126\cos 50^\circ.$

$\cos 50^\circ \approx 0.6428$ , so $a^2 \approx 130 - 80.99 = 49.01$ , giving $BC \approx 7.00\,\text{cm}$ .

Area $= \frac{1}{2}(7)(9)\sin 50^\circ = 31.5 \times 0.766 \approx 24.1\,\text{cm}^2$ .

Markers reward the correct cosine-rule setup, the value of $BC$ , the area formula $\frac{1}{2}ab\sin C$ , and the numerical area.

Related dot points

Sources & how we know this

CCEA GCE Mathematics specification — CCEA (2018)