Trigonometric proofs and methods of proof in QCE Specialist Mathematics Unit 3

What this dot point is asking

QCAA wants you to prove statements using the right logical structure, not just verify them on examples. Beyond induction you must handle direct proof, proof by contrapositive and proof by contradiction, and you must prove trigonometric identities by transforming one side into the other. The marks here reward a watertight chain of implications with justified steps, so sloppy logic loses credit even when the final claim is true.

The answer

Direct proof

A direct proof of "if $P$ then $Q$" assumes $P$ and derives $Q$ through a sequence of valid steps. For example, to prove that the product of two even integers is even, write $m = 2a$ and $n = 2b$ for integers $a, b$, so $mn = 4ab = 2(2ab)$, which is even because $2ab$ is an integer. Every step must follow from a definition or an earlier line.

Proof by contrapositive

The statement "if $P$ then $Q$" is logically equivalent to its contrapositive "if not $Q$ then not $P$". When the negation of $Q$ is easier to work with, prove the contrapositive instead. To prove "if $n^2$ is even then $n$ is even", the contrapositive "if $n$ is odd then $n^2$ is odd" is direct: $n = 2k+1$ gives $n^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$, which is odd.

Proof by contradiction

Assume the statement is false and derive a contradiction. The classic example is the irrationality of $\sqrt2$: assume $\sqrt2 = \frac{p}{q}$ in lowest terms, so $p^2 = 2q^2$. Then $p^2$ is even, so $p$ is even, say $p = 2r$, giving $4r^2 = 2q^2$, hence $q^2 = 2r^2$, so $q$ is also even. But then $p$ and $q$ share the factor $2$, contradicting "lowest terms". The assumption fails, so $\sqrt2$ is irrational.

Trigonometric proofs

To prove an identity such as $\frac{\sin 2\theta}{1 + \cos 2\theta} = \tan\theta$, work on one side using known identities until it matches the other. Standard tools include the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$, the double-angle formulas $\sin 2\theta = 2\sin\theta\cos\theta$ and $\cos 2\theta = 2\cos^2\theta - 1$, and the compound-angle formulas. Never move terms across the equals sign as if solving an equation; an identity proof transforms one expression, it does not manipulate both sides simultaneously.

Choosing a method

Direct proof suits constructive "build it forward" claims. Contrapositive helps when the negation of the conclusion is concrete (odd, nonzero, rational). Contradiction is the tool for non-existence and irrationality claims, where you assume the thing exists and break it. State which method you are using so the marker can follow the logic.

Writing valid logic

Each line must be justified: cite the definition, identity or earlier result used. Avoid the circular trap of assuming what you are proving. In an identity proof, declare a left-hand side (LHS) and right-hand side (RHS) and show LHS equals RHS by manipulating only one of them.

The identity toolkit

Most trigonometric proofs draw on a small, fixed set of identities, and fluency with them is what makes a proof flow. The Pythagorean family is $\sin^2\theta + \cos^2\theta = 1$, with the derived forms $1 + \tan^2\theta = \sec^2\theta$ and $1 + \cot^2\theta = \csc^2\theta$. The double-angle formulas are $\sin 2\theta = 2\sin\theta\cos\theta$ and $\cos 2\theta = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$, the three forms of $\cos 2\theta$ being chosen to match whatever the rest of the expression contains. The compound-angle formulas $\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B$ and $\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B$ generate the double-angle results when $A = B$ and handle sums of angles. A good first move on a stubborn identity is to convert everything to sines and cosines, since that exposes common factors to cancel.

Strategy: which side to start from

Begin from the more complicated side, because simplifying is easier than building complexity. If one side has a $\cos 2\theta$ or a product, expand it; if it has a fraction, look to combine or split it. When neither side is obviously simpler, transform both to a common third expression (such as everything in sines and cosines) and show each side equals that, which is still valid because you never operate across the equals sign. Reserve "if and only if" chains for equation solving, not identity proof.

Disproof by counterexample

Not every plausible statement is true, and the correct response to a false universal claim is a single counterexample. To disprove "$\sin(A + B) = \sin A + \sin B$ for all $A, B$", take $A = B = \tfrac{\pi}{2}$: the left side is $\sin\pi = 0$ while the right side is $1 + 1 = 2$. One counterexample is a complete disproof, whereas no number of confirming examples ever proves a universal statement, which is the whole reason induction and direct proof exist.