The Pythagorean and quotient identities, the reciprocal functions, the compound and double angle formulae, the R form for a sin theta plus b cos theta, and solving trigonometric equations over an interval.

What this dot point is asking

OCR wants you to use the Pythagorean identity $\sin^2\theta + \cos^2\theta \equiv 1$ and the quotient identity $\tan\theta \equiv \tfrac{\sin\theta}{\cos\theta}$, work with the reciprocal functions $\sec$, $\csc$ and $\cot$, apply the compound angle formulae and the double angle formulae, express $a\sin\theta + b\cos\theta$ in the form $R\sin(\theta \pm \alpha)$ or $R\cos(\theta \pm \alpha)$, prove trigonometric identities, and solve trigonometric equations over a stated interval.

The answer

The core identities

Compound and double angle formulae

The three forms of $\cos 2A$ are all useful: choose the one that introduces only the function you want. The last form is what makes $\cos 2A$ integrable, since $\sin^2 A = \tfrac{1}{2}(1 - \cos 2A)$.

Solving equations over an interval

Solve a trigonometric equation by reducing it to a single function, finding the principal value, then using the symmetry of the graph to find every solution in the interval. Beware of intervals on a transformed argument such as $2x$ or $x + 30^\circ$: widen the interval for the argument before solving, then convert back.

The R form

Writing $a\sin\theta + b\cos\theta$ as a single sinusoid makes its maximum, minimum and solutions obvious.

Examples in context

Proving identities

To prove an identity, work on the more complicated side and reduce it to the other using the core identities. Never move terms across the $\equiv$ sign as if solving an equation.

Equations on a transformed argument

Try this

Q1. Express $\sin\theta + \sqrt{3}\cos\theta$ in the form $R\sin(\theta + \alpha)$. [3 marks]

• Cue. $R = \sqrt{1 + 3} = 2$, $\tan\alpha = \sqrt{3}$, so $\alpha = 60^\circ$, giving $2\sin(\theta + 60^\circ)$.

Q2. Solve $\tan x = 2\sin x$ for $0^\circ \le x \le 360^\circ$. [3 marks]

• Cue. $\tfrac{\sin x}{\cos x} = 2\sin x$ gives $\sin x(1 - 2\cos x) = 0$, so $x = 0^\circ, 180^\circ, 360^\circ$ or $\cos x = \tfrac{1}{2}$ giving $x = 60^\circ, 300^\circ$.