Harmonic form and trigonometric proofEdexcel A-Level Maths: Revision notes

An expression $a\cos\theta+b\sin\theta$ combines two waves of the same period into one. It can be written as a single sine or cosine, called harmonic form: $a\cos\theta+b\sin\theta=R\cos(\theta-\alpha)\quad\text{or}\quad R\sin(\theta+\alpha)$ where $R>0$ is the amplitude and $\alpha$ the phase shift (acute unless told otherwise). This is useful because a single trig function is easy to solve, and its maximum and minimum values are simply $\pm R$.

Expand the form you want with a compound angle formula and compare coefficients. For $R\cos(\theta-\alpha)=R\cos\theta\cos\alpha+R\sin\theta\sin\alpha$ the match with $a\cos\theta+b\sin\theta$ gives $R\cos\alpha=a$ and $R\sin\alpha=b$, so $R=\sqrt{a^2+b^2},\qquad\tan\alpha=\frac ba.$ The four forms:

• $a\cos\theta+b\sin\theta=R\cos(\theta-\alpha)$, with $R\cos\alpha=a$, $R\sin\alpha=b$
• $a\cos\theta-b\sin\theta=R\cos(\theta+\alpha)$, with $R\cos\alpha=a$, $R\sin\alpha=b$
• $a\sin\theta+b\cos\theta=R\sin(\theta+\alpha)$, with $R\cos\alpha=a$, $R\sin\alpha=b$
• $a\sin\theta-b\cos\theta=R\sin(\theta-\alpha)$, with $R\cos\alpha=a$, $R\sin\alpha=b$ Example: $7\cos\theta-24\sin\theta=25\cos(\theta+\alpha)$, $\tan\alpha=\frac{24}{7}$, $\alpha=73.74^\circ$.

Once written as $R\cos(\theta-\alpha)$:

• Maximum value $R$, when $\theta-\alpha=0^\circ$ (or $360^\circ$); minimum value $-R$, when $\theta-\alpha=180^\circ$.
• To solve $a\cos\theta+b\sin\theta=c$ use $\cos(\theta-\alpha)=\frac cR$, working out the range of $\theta-\alpha$ first. If $|c|>R$ there is no solution.
• Expressions like $\dfrac{k}{p+a\cos\theta+b\sin\theta}$ are greatest or least when the denominator is least or greatest.

Express $\sin x-\sqrt3\cos x$ as $R\sin(x-\alpha)$ with $x$ in radians. $R\sin(x-\alpha)=R\sin x\cos\alpha-R\cos x\sin\alpha$, so $R\cos\alpha=1$, $R\sin\alpha=\sqrt3$. $R=\sqrt{1+3}=2$, $\tan\alpha=\sqrt3$, $\alpha=\frac\pi3$. Solve $=1$: $\sin\left(x-\frac\pi3\right)=\frac12$, with $-\frac\pi3\le x-\frac\pi3<\frac{5\pi}3$, giving $x-\frac\pi3=\frac\pi6,\frac{5\pi}6$, so $x=\frac\pi2,\frac{7\pi}6$.

To prove a trigonometric identity, start from one side (normally the more complicated) and transform it step by step into the other, stating the identity used at each step. Never treat the identity as true and work on both sides. Useful tools:

• $\tan\theta=\frac{\sin\theta}{\cos\theta}$ and $\sin^2\theta+\cos^2\theta=1$
• the double angle formulae, e.g. $1+\cos2\theta=2\cos^2\theta$ and $1-\cos2\theta=2\sin^2\theta$
• the compound angle formulae, read in either direction. Example: $\cos x\cos2x+\sin x\sin2x=\cos(2x-x)=\cos x$. Example: $\dfrac{\sin2\theta}{1+\cos2\theta}=\dfrac{2\sin\theta\cos\theta}{2\cos^2\theta}=\tan\theta$.

A proved identity can then be used to simplify an equation. Once $\dfrac{\sin2\theta}{1+\cos2\theta}\equiv\tan\theta$ is proved, the equation $\dfrac{\sin2\theta}{1+\cos2\theta}=3\tan^2\theta$ becomes $\tan\theta=3\tan^2\theta$, so $\tan\theta(3\tan\theta-1)=0$. Factorise, never divide by $\tan\theta$: this keeps $\tan\theta=0$ and gives $\theta=0^\circ,18.4^\circ,180^\circ,198.4^\circ$.