Mathematics 9709 · Pure Mathematics 3

Compound angle formulae

$\sin(A+B)$ is not $\sin A+\sin B$ . You cannot just add the angles inside sin and cos.
Each answer mixes together the sines and cosines of both angles.
The six formulae (and the two sign traps in them) are used all through this chapter.

$\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B$ and $\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B$ . Two signs to lock in: $\cos$ flips the sign (a plus turns into a minus), and the $\tan$ bottom takes the opposite sign to the top.

The six formulae (mind the signs)

$\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B$

$\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B$

$\tan(A\pm B)=\dfrac{\tan A\pm\tan B}{1\mp\tan A\tan B}$

These are given on the formula sheet, so the marks are for copying the signs right, not for remembering them.
Read $\mp$ as the opposite of the top sign: in $\cos(A+B)$ the top sign is $+$ , so the formula uses $-$ .

Worked examples

Worked example

Exact value of $\sin105^\circ$

Split $105^\circ$ into two special angles: $105^\circ=60^\circ+45^\circ$ .
Expand with the $\sin(A+B)$ formula (the plus stays a plus): $\sin105^\circ=\sin60^\circ\cos45^\circ+\cos60^\circ\sin45^\circ$ .
Substitute exact surds and tidy: $=\tfrac{\sqrt3}{2}\cdot\tfrac{1}{\sqrt2}+\tfrac12\cdot\tfrac{1}{\sqrt2}=\dfrac{\sqrt3+1}{2\sqrt2}=\dfrac{\sqrt6+\sqrt2}{4}$ .

Answer

\sin105^\circ=\dfrac{\sqrt6+\sqrt2}{4}

Worked example

Spot the formula: $\cos42^\circ\cos12^\circ+\sin42^\circ\sin12^\circ$

Do not reach for a calculator. Spot the pattern. The shape $\cos A\cos B+\sin A\sin B$ is exactly the right side of $\cos(A-B)$ (the plus here means a minus angle):

$=\cos(42^\circ-12^\circ)=\cos30^\circ=\tfrac{\sqrt3}{2}$

Answer

\tfrac{\sqrt3}{2}

Worked example

Real question: show $\tan(\theta+60^\circ)=2\cot\theta$ becomes $\tan^2\theta+3\sqrt3\tan\theta-2=0$

9709/32 O/N 2020 Q4(a). Plan: expand with the $\tan$ compound formula. Then clear the fractions until only one ratio is left.

Expand the left side with $\tan(A+B)$ and use $\tan60^\circ=\sqrt3$ : $\tan(\theta+60^\circ)=\dfrac{\tan\theta+\sqrt3}{1-\sqrt3\tan\theta}$ .
Write $2\cot\theta=\dfrac{2}{\tan\theta}$ , then cross-multiply: $\tan\theta(\tan\theta+\sqrt3)=2(1-\sqrt3\tan\theta)$ .
Expand both sides: $\tan^2\theta+\sqrt3\tan\theta=2-2\sqrt3\tan\theta$ .
Collect the $\tan\theta$ terms ( $\sqrt3+2\sqrt3=3\sqrt3$ ): $\tan^2\theta+3\sqrt3\tan\theta-2=0$ .

Answer

\tan^2\theta+3\sqrt3\tan\theta-2=0\ \checkmark

Why a sum of two waves can shift sideways

What the compound formula is saying

Read $\sin(A+B)=\sin A\cos B+\cos A\sin B$ with $B$ held fixed. It says $\sin(A+B)$ is a fixed mix of $\sin A$ and $\cos A$ , using fixed amounts $\cos B$ and $\sin B$ of each. Run that backwards and you reach the point of §3.5. Any mix $a\sin A+b\cos A$ can be rewritten as one shifted sine $R\sin(A+\alpha)$ . The compound formula proves it can be done.

Where the marks go

Common mistake

\cos(A+B)=\cos A\cos B-\sin A\sin B

: a minus, even though the angle is a sum. The sin formula keeps the plus; the cos formula flips it.

Common mistake

The sign on the

\tan

bottom is the opposite of the top:

\tan(A+B)

has

1-\tan A\tan B

underneath. (9709/32 Oct/Nov 2020 Q4)

Now you try

Without a calculator, find the exact value of $\tan75^\circ$ . (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

$\tan(45^\circ+30^\circ)=\dfrac{1+\tfrac{1}{\sqrt3}}{1-\tfrac{1}{\sqrt3}}=\dfrac{\sqrt3+1}{\sqrt3-1}$ ; rationalise by multiplying top and bottom by $\sqrt3+1$ :

Answer

\tan75^\circ=2+\sqrt3

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Compound angle formulae

$\sin(A+B)$ is not $\sin A+\sin B$ . You cannot just add the angles inside sin and cos.
Each answer mixes together the sines and cosines of both angles.
The six formulae (and the two sign traps in them) are used all through this chapter.

The six formulae (mind the signs)

$\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B$

$\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B$

$\tan(A\pm B)=\dfrac{\tan A\pm\tan B}{1\mp\tan A\tan B}$

These are given on the formula sheet, so the marks are for copying the signs right, not for remembering them.
Read $\mp$ as the opposite of the top sign: in $\cos(A+B)$ the top sign is $+$ , so the formula uses $-$ .

Worked examples

Worked example

Exact value of $\sin105^\circ$

Split $105^\circ$ into two special angles: $105^\circ=60^\circ+45^\circ$ .
Expand with the $\sin(A+B)$ formula (the plus stays a plus): $\sin105^\circ=\sin60^\circ\cos45^\circ+\cos60^\circ\sin45^\circ$ .
Substitute exact surds and tidy: $=\tfrac{\sqrt3}{2}\cdot\tfrac{1}{\sqrt2}+\tfrac12\cdot\tfrac{1}{\sqrt2}=\dfrac{\sqrt3+1}{2\sqrt2}=\dfrac{\sqrt6+\sqrt2}{4}$ .

Answer

\sin105^\circ=\dfrac{\sqrt6+\sqrt2}{4}

Worked example

Spot the formula: $\cos42^\circ\cos12^\circ+\sin42^\circ\sin12^\circ$

Do not reach for a calculator. Spot the pattern. The shape $\cos A\cos B+\sin A\sin B$ is exactly the right side of $\cos(A-B)$ (the plus here means a minus angle):

$=\cos(42^\circ-12^\circ)=\cos30^\circ=\tfrac{\sqrt3}{2}$

Answer

\tfrac{\sqrt3}{2}

Worked example

Real question: show $\tan(\theta+60^\circ)=2\cot\theta$ becomes $\tan^2\theta+3\sqrt3\tan\theta-2=0$

9709/32 O/N 2020 Q4(a). Plan: expand with the $\tan$ compound formula. Then clear the fractions until only one ratio is left.

Expand the left side with $\tan(A+B)$ and use $\tan60^\circ=\sqrt3$ : $\tan(\theta+60^\circ)=\dfrac{\tan\theta+\sqrt3}{1-\sqrt3\tan\theta}$ .
Write $2\cot\theta=\dfrac{2}{\tan\theta}$ , then cross-multiply: $\tan\theta(\tan\theta+\sqrt3)=2(1-\sqrt3\tan\theta)$ .
Expand both sides: $\tan^2\theta+\sqrt3\tan\theta=2-2\sqrt3\tan\theta$ .
Collect the $\tan\theta$ terms ( $\sqrt3+2\sqrt3=3\sqrt3$ ): $\tan^2\theta+3\sqrt3\tan\theta-2=0$ .

Answer

\tan^2\theta+3\sqrt3\tan\theta-2=0\ \checkmark

Why a sum of two waves can shift sideways

What the compound formula is saying

Where the marks go

Common mistake

\cos(A+B)=\cos A\cos B-\sin A\sin B

: a minus, even though the angle is a sum. The sin formula keeps the plus; the cos formula flips it.

Common mistake

The sign on the

\tan

bottom is the opposite of the top:

\tan(A+B)

has

1-\tan A\tan B

underneath. (9709/32 Oct/Nov 2020 Q4)

Now you try

Without a calculator, find the exact value of $\tan75^\circ$ . (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

$\tan(45^\circ+30^\circ)=\dfrac{1+\tfrac{1}{\sqrt3}}{1-\tfrac{1}{\sqrt3}}=\dfrac{\sqrt3+1}{\sqrt3-1}$ ; rationalise by multiplying top and bottom by $\sqrt3+1$ :

Answer

\tan75^\circ=2+\sqrt3