Addition and double angle formulaeEdexcel International A Level Maths: Revision notes

The formulae for the sine, cosine and tangent of a sum or difference are: $\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B$ $\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B$ $\tan(A\pm B)=\frac{\tan A\pm\tan B}{1\mp\tan A\tan B}$ Notice that the sign in $\sin(A\pm B)$ is the same as in the bracket, but the sign in $\cos(A\pm B)$ is reversed. Learn all of them, including the double angle formulae.

Write an unfamiliar angle as a sum or difference of angles with known exact values. Example. $\sin75^{\circ}=\sin(45^{\circ}+30^{\circ})=\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2}+\frac{\sqrt2}{2}\cdot\frac12=\frac{\sqrt6+\sqrt2}{4}$. Likewise $\cos75^{\circ}=\frac{\sqrt6-\sqrt2}{4}$, and dividing gives $\tan75^{\circ}=\frac{\sqrt6+\sqrt2}{\sqrt6-\sqrt2}=2+\sqrt3$ after rationalising the denominator.

Putting $B=A$ in the addition formulae gives the double angle formulae: $\sin2A=2\sin A\cos A,$ $\cos2A=\cos^2A-\sin^2A=2\cos^2A-1=1-2\sin^2A,$ $\tan2A=\frac{2\tan A}{1-\tan^2A}.$ The second and third forms of $\cos2A$ come from $\sin^2A+\cos^2A=1$. Choose the form that suits the question: $2\cos^2A-1$ when the answer involves cosine only.

Any double angle formula can be read with $A=\frac{\theta}{2}$. For example $\cos\theta=2\cos^2\frac{\theta}{2}-1=1-2\sin^2\frac{\theta}{2}$, and $\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$. If $\theta$ is acute and $\cos\theta=\frac{7}{25}$, then $\cos^2\frac{\theta}{2}=\frac{1+7/25}{2}=\frac{16}{25}$, so $\cos\frac{\theta}{2}=\frac45$ (positive because $\frac{\theta}{2}$ is acute) and $\sin\frac{\theta}{2}=\frac35$. Always decide the sign of a square root from the quadrant of the half angle.

Proving. Start from the more complicated side and reduce it. To prove $\cos x\cos2x+\sin x\sin2x\equiv\cos x$, recognise $\cos A\cos B+\sin A\sin B=\cos(A-B)$ with $A=2x$ and $B=x$: the left side is $\cos(2x-x)=\cos x$. Triple angle. $\cos3x=\cos(2x+x)=(2\cos^2x-1)\cos x-2\sin^2x\cos x=4\cos^3x-3\cos x$. Solving. Replace $\sin2x$ or $\cos2x$ so the equation involves one function of $x$, then factorise. For $\sin2x=\sin x$: $2\sin x\cos x-\sin x=0$, so $\sin x(2\cos x-1)=0$ and $x=0^{\circ},60^{\circ},180^{\circ},300^{\circ}$.