Learn the method, study an example, then explain your reasoning in an exam-style problem. Work on paper before opening the solution and compare each step with the marking guidance.
Identities, factorisation and valid domains
Transform trigonometric expressions while retaining every domain restriction.
Notes
Before you start: simplify fractions and factor a product without losing a zero factor.
The core relations are \(\sin^2x+\cos^2x=1\) for real x and \(\tan x=\sin x/\cos x\) where \(\cos x\ne0\). Dividing the first identity by cos²x gives \(1+\tan^2x=1/\cos^2x\) only on that same domain.
In a proof, start with one side and transform it into the other using valid equalities. An identity holds at every point of its stated domain; an equation may hold only at selected angles. A cancellation can simplify a formula without filling a hole in the original domain.
If sin x or cos x is a factor in an equation, check its zero case before division. Multiplying by a denominator or squaring can admit extra candidates, so check solutions in the original expression.
Recall later
\((1-\cos^2x)/\sin x=\sin x\) when \(\sin x\ne0\); the original quotient is still undefined at x=0.
Common mistakes
Cancelling a factor without retaining its excluded values.
Dividing an equation by sine or cosine and losing roots.
Using a chain of implications in place of valid equalities in an identity proof.
Three worked examples
Example 1
Simplify \((1-\sin^2x)/\cos x\) and state its restriction.
Show the reasoning
Replace the numerator by cos²x and cancel one cosine: the result is cos x, with the original restriction \(\cos x\ne0\), or \(x\ne\pi/2+n\pi\).
Example 2
Show that \(\sin^2x+2\cos^2x=1+\cos^2x\).
Show the reasoning
Group one sin²x+cos²x pair: \((\sin^2x+\cos^2x)+\cos^2x=1+\cos^2x\). No division is used, so this identity is valid for every real x.
Example 3
Solve \(2\sin x\cos x=\sin x\) for \(0\le x<2\pi\).
Show the reasoning
Factor \(\sin x(2\cos x-1)=0\). The first factor gives x=0,π; the second gives x=π/3,5π/3. Dividing by sine would lose the first two solutions.
