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

The compound angle formulae give the sine, cosine and tangent of a sum or difference of two angles: $\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B$ $\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B$ $\tan(A\pm B)=\frac{\tan A\pm\tan B}{1\mp\tan A\tan B}$ Note the signs: sine keeps the sign of the bracket, cosine reverses it, and in $\tan$ the denominator reverses it. Learn them so you can apply them quickly. Example: if $\sin A=\frac35$, $\cos B=\frac5{13}$ ($A,B$ acute) then $\cos A=\frac45$, $\sin B=\frac{12}{13}$ and $\sin(A+B)=\frac{15+48}{65}=\frac{63}{65}$.

You need to understand a geometrical proof of the sine and cosine formulae. On the unit circle the point at angle $A+B$ has height $\sin(A+B)$. Build it from two right-angled triangles. The first has hypotenuse $1$ and angle $B$, with sides $\cos B$ and $\sin B$. Use each of those sides as the hypotenuse of a second right-angled triangle with angle $A$ (the one on the $\sin B$ side is rotated by $A$). Adding the vertical heights gives $\sin(A+B)=\sin A\cos B+\cos A\sin B$, and combining the horizontal lengths (one is subtracted) gives $\cos(A+B)=\cos A\cos B-\sin A\sin B$. Replacing $B$ by $-B$, with $\cos(-B)=\cos B$ and $\sin(-B)=-\sin B$, gives the formulae for $A-B$. Dividing sine by cosine gives the tangent formula.

Put $B=A$ in the compound formulae: $\sin2A=2\sin A\cos A$ $\cos2A=\cos^2A-\sin^2A=2\cos^2A-1=1-2\sin^2A$ $\tan2A=\frac{2\tan A}{1-\tan^2A}$ The three forms of $\cos2A$ come from $\sin^2A+\cos^2A=1$; choose the one that leaves only the ratio you have. The angle can be anything: $\cos4x=2\cos^22x-1$, and $\sin\theta=2\sin\frac\theta2\cos\frac\theta2$. Example: $\tan x=\frac34$ gives $\tan2x=\frac{3/2}{7/16}=\frac{24}7$ and $\cos2x=\frac{1-\tan^2x}{1+\tan^2x}=\frac7{25}$.

Write an awkward angle as a sum or difference of angles whose exact values you know ($30^\circ,45^\circ,60^\circ$). Example: $\sin75^\circ=\sin(45^\circ+30^\circ)=\frac{\sqrt2}2\cdot\frac{\sqrt3}2+\frac{\sqrt2}2\cdot\frac12=\frac{\sqrt6+\sqrt2}4$. Similarly $\cos75^\circ=\frac{\sqrt6-\sqrt2}4$, so $\tan75^\circ=\frac{\sqrt3+1}{\sqrt3-1}=2+\sqrt3$ after rationalising. To prove an identity, work on one side only (usually the more complicated) until it matches the other, citing the formula used at each step.

Replace the double angle so that the equation involves a single trig function, then factorise or use a quadratic. Example: $\cos2\theta+3\sin\theta=2$ becomes $1-2\sin^2\theta+3\sin\theta=2$, so $2\sin^2\theta-3\sin\theta+1=0$, i.e. $(2\sin\theta-1)(\sin\theta-1)=0$. Then $\sin\theta=\frac12$ or $1$, giving $\theta=30^\circ,150^\circ,90^\circ$ in $[0,360^\circ)$. For $\sin2x=k$ solve for $2x$ over the doubled interval first (for $0\le x<360^\circ$ use $0\le2x<720^\circ$), then halve.

Write the left-hand side as a single cosine: $a\cos\theta+b\sin\theta=R\cos(\theta-\alpha)$, where $R=\sqrt{a^2+b^2}$, $R\cos\alpha=a$ and $R\sin\alpha=b$. Then solve $R\cos(\theta-\alpha)=c$. Example: $\sqrt3\cos\theta+\sin\theta=1$: $R=2$, $\alpha=30^\circ$, so $\cos(\theta-30^\circ)=\frac12$, $\theta-30^\circ=\pm60^\circ$, $\theta=90^\circ$ or $330^\circ$. Check the interval for $\theta-\alpha$ before solving, and check by substituting back. The maximum value of the expression is $R$, when $\theta=\alpha$.