Skip to main content
Northern IrelandFurther MathsSyllabus dot point

How do you solve trigonometric equations and use the sine rule, cosine rule and trigonometric identities?

Use trigonometry: apply the sine rule, cosine rule and area formula, use the identities for tan and the Pythagorean identity, and solve trigonometric equations over a given interval.

A CCEA GCSE Further Mathematics answer on trigonometry, covering the sine and cosine rules, the area formula, exact values, the tan and Pythagorean identities, and solving trigonometric equations across an interval in the compulsory Pure unit.

Generated by Claude Opus 4.812 min answer

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

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this dot point is asking
  2. The sine and cosine rules
  3. Exact values and the unit-circle behaviour
  4. Trigonometric identities
  5. Solving trigonometric equations
  6. Why this matters

What this dot point is asking

Trigonometry in CCEA GCSE Further Mathematics extends the right-angled work of ordinary GCSE to any triangle and to equations. You must use the sine rule, the cosine rule and the area formula for non-right-angled triangles, know and use the tan identity and the Pythagorean identity to simplify and prove, and solve trigonometric equations across a stated interval, finding every solution. These skills support the coordinate-geometry and calculus topics and appear on both calculator and exact-value questions.

The sine and cosine rules

The sine and cosine rules handle triangles that are not right-angled. Use the sine rule when you have a matching side and angle pair; use the cosine rule when you have two sides and the included angle, or all three sides.

Choosing the right rule is half the marks. If the angle sits between the two sides you know, the cosine rule finds the third side; if you can pair a side with its opposite angle, the sine rule is quicker.

Exact values and the unit-circle behaviour

You should know the exact values of sine, cosine and tangent for the special angles 0,30,45,600^\circ, 30^\circ, 45^\circ, 60^\circ and 9090^\circ, such as sin30=12\sin 30^\circ = \tfrac{1}{2}, cos30=32\cos 30^\circ = \tfrac{\sqrt{3}}{2} and tan45=1\tan 45^\circ = 1. You also need the signs of each ratio in each quadrant, which determine where extra solutions to an equation lie: sine is positive for 00^\circ to 180180^\circ, cosine is positive for 00^\circ to 9090^\circ and 270270^\circ to 360360^\circ, and tangent repeats every 180180^\circ.

Trigonometric identities

The two identities let you rewrite expressions and prove results. The tan identity turns a ratio of sine and cosine into a tangent, and the Pythagorean identity links the squares of sine and cosine.

These are used both to simplify (replacing 1cos2θ1 - \cos^2\theta by sin2θ\sin^2\theta) and to convert an equation into one in a single ratio so it can be solved.

Solving trigonometric equations

A trigonometric equation usually has several solutions in an interval, because the graphs repeat. Find the first value, then use the graph's symmetry to find the rest.

Why this matters

Trigonometry is essential across the qualification. The sine and cosine rules solve real geometry and feed the Mechanics unit through resolving forces and velocities; the identities streamline proofs and equation solving; and exact values keep non-calculator answers tidy. Solving equations correctly over an interval, finding every root, is a frequently tested and frequently lost set of marks.

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 Unit 1 (style)4 marksSolve 2sinx=12\sin x = 1 for 0x3600^\circ \le x \le 360^\circ.
Show worked answer →

Rearrange: sinx=12\sin x = \tfrac{1}{2}.

The first solution is x=sin1(0.5)=30x = \sin^{-1}(0.5) = 30^\circ.

Sine is also positive in the second quadrant, so the other solution is 18030=150180^\circ - 30^\circ = 150^\circ.

Therefore x=30x = 30^\circ or x=150x = 150^\circ. Marks are for the principal value and for the second solution in range. Forgetting the second value is the most common error in interval questions.

CCEA Unit 1 (style)5 marksIn triangle ABCABC, AB=7AB = 7 cm, AC=9AC = 9 cm and angle BAC=58BAC = 58^\circ. Find BCBC and the area of the triangle.
Show worked answer →

Use the cosine rule for the side opposite the known angle: BC2=72+922(7)(9)cos58BC^2 = 7^2 + 9^2 - 2(7)(9)\cos 58^\circ.

BC2=49+81126cos58=130126(0.5299)=13066.77=63.23BC^2 = 49 + 81 - 126\cos 58^\circ = 130 - 126(0.5299) = 130 - 66.77 = 63.23, so BC=7.95BC = 7.95 cm.

Area =12absinC=12(7)(9)sin58=31.5×0.8480=26.7= \tfrac{1}{2}ab\sin C = \tfrac{1}{2}(7)(9)\sin 58^\circ = 31.5 \times 0.8480 = 26.7 cm2^2.

Marks are for the cosine rule, the side, the area formula, and the area. A common slip is using the sine rule when the angle is between the two known sides.

Related dot points

Sources & how we know this