Compound and double angle formulaeAQA A-Level Maths: Revision notes

The compound angle formulae expand a trig function of $A\pm B$: $\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B$ $\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B$ $\tan(A\pm B)=\frac{\tan A\pm\tan B}{1\mp\tan A\tan B}.$ Notice that in $\cos(A\pm B)$ the sign changes, and in the denominator of $\tan(A\pm B)$ it changes too. They are not linear: $\sin(A+B)\ne\sin A+\sin B$. Example: with $\sin A=\frac35$, $\cos B=\frac5{13}$ (both acute), $\cos A=\frac45$ and $\sin B=\frac{12}{13}$, so $\cos(A+B)=\frac{4}{5}\cdot\frac{5}{13}-\frac35\cdot\frac{12}{13}=-\frac{16}{65}$.

Put $B=A$ in the compound formulae: $\sin2A=2\sin A\cos A$ $\cos2A=\cos^2A-\sin^2A=2\cos^2A-1=1-2\sin^2A$ $\tan2A=\frac{2\tan A}{1-\tan^2A}.$ There are three forms of $\cos2A$: choose the one that leaves a single trig function in your equation. Rearranged, $\cos^2A=\frac{1+\cos2A}{2}$ and $\sin^2A=\frac{1-\cos2A}{2}$. Example: if $\tan\theta=\frac34$, then $\sin\theta=\frac35$, $\cos\theta=\frac45$, and $\sin2\theta=\frac{24}{25}$, $\cos2\theta=\frac{7}{25}$, $\tan2\theta=\frac{24}{7}$.

Splitting an angle into two with known values gives exact results. Example: $75^\circ=45^\circ+30^\circ$: $\sin75^\circ=\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2}+\frac{\sqrt2}{2}\cdot\frac12=\frac{\sqrt6+\sqrt2}{4},\qquad\cos75^\circ=\frac{\sqrt6-\sqrt2}{4}.$ So $\tan75^\circ=\frac{\sqrt6+\sqrt2}{\sqrt6-\sqrt2}$, and multiplying top and bottom by $\sqrt6+\sqrt2$ gives $\frac{8+4\sqrt3}{4}=2+\sqrt3$. Learn the exact values for $30^\circ$, $45^\circ$ and $60^\circ$ first.

Replace the double angle with the form that makes a quadratic in one function, factorise, and list every solution in the interval. Example: $\cos2\theta+3\sin\theta=2$ becomes $2\sin^2\theta-3\sin\theta+1=0$, so $\sin\theta=\frac12$ or $1$, giving $\theta=\frac{\pi}{6},\frac{5\pi}{6},\frac{\pi}{2}$. If the equation is $\sin2\theta=\sin\theta$, write $2\sin\theta\cos\theta-\sin\theta=0$ and factorise: never divide by $\sin\theta$, or you lose the solutions where $\sin\theta=0$.

To prove an identity, start from one side, apply the formulae and simplify. Example: $\frac{\sin2\theta}{1+\cos2\theta}=\frac{2\sin\theta\cos\theta}{2\cos^2\theta}=\tan\theta$, using $1+\cos2\theta=2\cos^2\theta$. Geometrical proof of $\sin(A+B)$: draw a line of length $h$ from vertex $P$ to the base, splitting the angle at $P$ into $A$ and $B$ and meeting the base at right angles. The two sides from $P$ are $\frac{h}{\cos A}$ and $\frac{h}{\cos B}$, and the base is split into $h\tan A$ and $h\tan B$. The area of the whole triangle is $\frac12\cdot\frac{h}{\cos A}\cdot\frac{h}{\cos B}\sin(A+B)$, and it also equals $\frac12h(h\tan A+h\tan B)$. Equating and multiplying by $\cos A\cos B$ gives $\sin(A+B)=\sin A\cos B+\cos A\sin B$.