The forms R cos(theta ± a) and R sin(theta ± a)Edexcel International A Level Maths: Revision notes

An expression $a\cos\theta+b\sin\theta$ is the sum of two waves with the same period. It can be written as a single wave $R\cos(\theta\pm\alpha)$ or $R\sin(\theta\pm\alpha)$ (the specification writes $r$ for $R$). This makes the maximum, minimum and solutions of equations easy to find. Usually $R>0$ and $\alpha$ is acute. Expand the target form using the addition formulae:

• $R\cos(\theta+\alpha)=R\cos\alpha\cos\theta-R\sin\alpha\sin\theta$
• $R\cos(\theta-\alpha)=R\cos\alpha\cos\theta+R\sin\alpha\sin\theta$
• $R\sin(\theta+\alpha)=R\cos\alpha\sin\theta+R\sin\alpha\cos\theta$
• $R\sin(\theta-\alpha)=R\cos\alpha\sin\theta-R\sin\alpha\cos\theta$

Compare coefficients of $\cos\theta$ and $\sin\theta$ in the expansion. This gives two equations, for $R\cos\alpha$ and $R\sin\alpha$.

• Square and add: $R^2(\cos^2\alpha+\sin^2\alpha)=R^2$, so $R=\sqrt{a^2+b^2}$.
• Divide: $\tan\alpha=\frac{R\sin\alpha}{R\cos\alpha}$. Example. Write $3\sin\theta+4\cos\theta$ as $R\sin(\theta+\alpha)$. Then $R\cos\alpha=3$ and $R\sin\alpha=4$, so $R=5$ and $\tan\alpha=\frac43$, giving $\alpha=53.1^{\circ}$. Example. $\cos\theta-\sqrt3\sin\theta=R\cos(\theta+\alpha)$ gives $R\cos\alpha=1$ and $R\sin\alpha=\sqrt3$, so $R=2$ and $\alpha=\frac{\pi}{3}$.

The wave $R\sin(\theta+\alpha)$ lies between $-R$ and $R$. Its greatest value $R$ occurs when $\sin(\theta+\alpha)=1$, that is $\theta+\alpha=90^{\circ}$ (or $\frac{\pi}{2}$), and its least value $-R$ when $\sin(\theta+\alpha)=-1$. So $3\sin\theta+4\cos\theta$ has maximum $5$ at $\theta=90^{\circ}-53.1^{\circ}=36.9^{\circ}$, and $\cos\theta-\sqrt3\sin\theta=2\cos\left(\theta+\frac{\pi}{3}\right)$ has minimum $-2$ when $\theta+\frac{\pi}{3}=\pi$, so $\theta=\frac{2\pi}{3}$. For a model such as $D=6+2\cos x+3\sin x$, the extremes of $D$ are $6\pm\sqrt{13}$.

Rewrite the left side as a single wave, then solve like a basic trigonometric equation. Example. Solve $5\cos\theta-12\sin\theta=6$ for $0\le\theta<360^{\circ}$. Write it as $13\cos(\theta+67.4^{\circ})=6$, so $\cos(\theta+67.4^{\circ})=\frac{6}{13}$. The interval for $\theta+67.4^{\circ}$ is $67.4^{\circ}\le\theta+67.4^{\circ}<427.4^{\circ}$. $\cos^{-1}\frac{6}{13}=62.5^{\circ}$ is below the interval, so use $360^{\circ}-62.5^{\circ}=297.5^{\circ}$ and $360^{\circ}+62.5^{\circ}=422.5^{\circ}$. Then $\theta=230.1^{\circ}$ or $355.1^{\circ}$. If $|c|>R$ there are no solutions, because the wave never reaches $c$.

Modelling questions (tides, daylight, alternating current) often write a sum of a sine and a cosine of the same angle. Convert it to $R\sin(x+\alpha)$ to read off the maximum and minimum, and to find when the quantity is above a given level. For the depth $D=6+2\cos x+3\sin x$ with $x=\frac{\pi t}{6}$: $D=6+\sqrt{13}\sin(x+0.588)$. The maximum is $9.61$ m at $t=1.88$ h. The depth is at least $9$ m when $\sin(x+0.588)\ge\frac{3}{\sqrt{13}}$, which gives $0.754\le t\le3$.