A-Maths Trigonometric Proofs: Choose a Step That Moves Towards the Target

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A Secondary 4 student in a public SGExams discussion on trigonometric proving could complete basic proofs but struggled to choose the next step in harder ones. Solutions looked straightforward afterwards. That gap suggests a useful practice goal: explain why a particular transformation helps, rather than only remembering the finished working.

The official 2026 Additional Mathematics syllabus includes simplifying trigonometric expressions and proving simple identities. The original examples below demonstrate a way to plan those transformations; they are not predictions of examination questions.

Before writing, compare the two sides. Does one contain a double angle while the other contains single angles? Does one have a fraction and the other a product? Are several trigonometric functions being reduced to just one? These differences suggest what needs to change. Starting with the more complicated side often helps, but the useful side is the one with a clear route towards the target.

Example one: recognise a difference of squares

Prove that cos x ÷ (1 − sin x) = (1 + sin x) ÷ cos x, wherever both sides are defined. The target contains 1 + sin x, while the starting denominator contains 1 − sin x. That pairing suggests multiplying by 1 + sin x.

Starting from the left-hand side, multiply numerator and denominator by 1 + sin x. This gives cos x(1 + sin x) ÷ [(1 − sin x)(1 + sin x)]. The denominator becomes 1 − sin²x, which equals cos²x. Cancelling a common factor cos x then gives (1 + sin x) ÷ cos x, the required right-hand side.

The cancellation concerns a factor of the entire numerator and denominator. It is not cancellation across addition. Keep the brackets until the factorisation is clear. The common domain excludes cos x = 0; on that domain the transformations used here are valid.

Example two: match the angle first

Consider (1 − cos 2x) ÷ sin 2x = tan x, wherever both sides are defined. The left contains double angles, while the target is a single-angle ratio. Use 1 − cos 2x = 2sin²x and sin 2x = 2sin x cos x. The fraction becomes 2sin²x ÷ (2sin x cos x), which simplifies to sin x ÷ cos x = tan x.

Here the original denominator requires sin 2x to be non-zero. That condition means both sin x and cos x are non-zero, so the cancellation is legitimate. The simplified expression tan x is defined at some additional points, but those points do not become part of the original identity’s domain.

The reason for choosing the double-angle identities was visible before calculating: they turn the left-hand side into the sine-over-cosine structure of the target. This is more useful than applying an identity simply because it contains a familiar symbol.

Keep exploration separate from the proof

On rough paper, you can explore both sides, try a substitution, or work backwards from the target. In the final solution, present a valid chain of equal expressions from one side to the other. Avoid assuming the desired equality, rearranging it and treating a resulting true statement as a complete proof unless every implication has been justified.

A numerical check is useful for catching errors. If the two sides disagree at an allowed angle, something is wrong. Agreement at several angles does not prove an identity for every allowed angle. The written algebra supplies that proof.

Turn each solution into a reusable decision

After checking a question, annotate the decisive step with a short reason: “combine fractions to create one numerator”, “replace double angles to match tan x”, or “factorise before cancelling”. Then close the solution and reproduce that step from the original expression.

Choose a few appropriate questions from the TestPapersFree A-Maths collection. Mix their structures rather than repeating only one pattern. If you get stuck, record the exact expression where progress stopped and ask your teacher which feature should have guided the next move. That makes a consultation much more specific than asking for general trigonometry tips.