Compound angle and double angle formulae

Compound angle formulae express a trigonometric function of a sum or difference, such as $\sin(A+B)$, in terms of functions of $A$ and $B$. They generate exact values, double angle identities, proofs and methods for solving equations.

The three core formulae are

$$\boxed{\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B},$$ $$\boxed{\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B},$$

and

$$\boxed{\tan(A\pm B)=\frac{\tan A\pm\tan B}{1\mp\tan A\tan B}},$$

where the tangent expression is used only when all the quantities involved are defined.

Prerequisites

You should be able to:

• recall exact values of sine, cosine and tangent at standard angles;
• use radians as well as degrees;
• simplify algebraic fractions and surds;
• use $\sin^2x+\cos^2x=1$;
• solve basic trigonometric equations over a stated interval.

Review exact trigonometric values, radians or trigonometric equations where needed.

Reading the formulae accurately

The notation $A\pm B$ represents two separate statements. For example,

$$\sin(A+B)=\sin A\cos B+\cos A\sin B,$$ $$\sin(A-B)=\sin A\cos B-\cos A\sin B.$$

For sine, the sign in the expansion is the same as the sign between the angles. For cosine and tangent, the displayed second sign changes:

Function	Sum	Difference
sine	$\sin A\cos B+\cos A\sin B$	$\sin A\cos B-\cos A\sin B$
cosine	$\cos A\cos B-\sin A\sin B$	$\cos A\cos B+\sin A\sin B$
tangent	$\dfrac{\tan A+\tan B}{1-\tan A\tan B}$	$\dfrac{\tan A-\tan B}{1+\tan A\tan B}$

There is no distributive rule

$\sin(A+B)$ is not $\sin A+\sin B$, and $\cos(A+B)$ is not $\cos A+\cos B$. For instance,

$$\sin(30^\circ+60^\circ)=1,$$

but $\sin30^\circ+\sin60^\circ=(1+\sqrt3)/2\ne1$.

Worked example 1: expand a difference

Expand $\cos(3x-20^\circ)$.

Use $A=3x$ and $B=20^\circ$. Cosine of a difference has a plus sign between its two products:

$$\boxed{\cos(3x-20^\circ) =\cos3x\cos20^\circ+\sin3x\sin20^\circ.}$$

The angle $3x$ must remain intact. Writing $3\cos x$ or $\cos3\cos x$ would be invalid.

Self check 1

Expand $\sin(2x+\alpha)$ and $\tan(p-q)$.

Answer

$$\sin(2x+\alpha)=\sin2x\cos\alpha+\cos2x\sin\alpha,$$ $$\tan(p-q)=\frac{\tan p-\tan q}{1+\tan p\tan q}.$$

Exact values at non standard angles

Write the target angle as a sum or difference of standard angles. Choose a decomposition for which every required value is known exactly.

Worked example 2: find $\sin75^\circ$

Since $75^\circ=45^\circ+30^\circ$,

$$\begin{aligned} \sin75^\circ &=\sin(45^\circ+30^\circ)\\ &=\sin45^\circ\cos30^\circ+\cos45^\circ\sin30^\circ\\ &=\frac{1}{\sqrt2}\cdot\frac{\sqrt3}{2} +\frac{1}{\sqrt2}\cdot\frac12\\ &=\boxed{\frac{\sqrt6+\sqrt2}{4}}. \end{aligned}$$

This is plausible because $75^\circ$ is in the first quadrant and $\sin75^\circ$ is close to $1$.

Worked example 3: find $\tan15^\circ$

Use $15^\circ=45^\circ-30^\circ$:

$$\begin{aligned} \tan15^\circ &=\frac{\tan45^\circ-\tan30^\circ} {1+\tan45^\circ\tan30^\circ}\\ &=\frac{1-1/\sqrt3}{1+1/\sqrt3}\\ &=\frac{\sqrt3-1}{\sqrt3+1}. \end{aligned}$$

Rationalising gives

$$\tan15^\circ =\frac{(\sqrt3-1)^2}{3-1} =\frac{4-2\sqrt3}{2} =\boxed{2-\sqrt3}.$$

Worked example 4: an angle in radians

Find the exact value of $\cos\dfrac{\pi}{12}$.

Because $\dfrac{\pi}{12}=\dfrac{\pi}{4}-\dfrac{\pi}{6}$,

$$\begin{aligned} \cos\frac{\pi}{12} &=\cos\left(\frac{\pi}{4}-\frac{\pi}{6}\right)\\ &=\cos\frac{\pi}{4}\cos\frac{\pi}{6} +\sin\frac{\pi}{4}\sin\frac{\pi}{6}\\ &=\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2} +\frac{\sqrt2}{2}\cdot\frac12\\ &=\boxed{\frac{\sqrt6+\sqrt2}{4}}. \end{aligned}$$

Radians do not change the identity. They change only how angles are written.

Self check 2

Find exact values of:

1. $\cos105^\circ$;
2. $\sin\dfrac{\pi}{12}$.

Answer

Using $105^\circ=60^\circ+45^\circ$,

$$\cos105^\circ =\frac12\cdot\frac{\sqrt2}{2} -\frac{\sqrt3}{2}\cdot\frac{\sqrt2}{2} =\boxed{\frac{\sqrt2-\sqrt6}{4}}.$$

The negative sign agrees with the second quadrant.

Using $\pi/12=\pi/4-\pi/6$,

$$\sin\frac{\pi}{12} =\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2} -\frac{\sqrt2}{2}\cdot\frac12 =\boxed{\frac{\sqrt6-\sqrt2}{4}}.$$

Where the tangent formula comes from

The tangent formula need not be memorised independently. Start from $\tan\theta=\sin\theta/\cos\theta$:

$$\begin{aligned} \tan(A+B) &=\frac{\sin(A+B)}{\cos(A+B)}\\ &=\frac{\sin A\cos B+\cos A\sin B} {\cos A\cos B-\sin A\sin B}. \end{aligned}$$

Divide numerator and denominator by $\cos A\cos B$:

$$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}.$$

This derivation also exposes a domain issue. If $\cos A\cos B=0$, that division is not allowed. More generally, never use the tangent formula blindly at angles where a tangent is undefined.

Worked example 5: recover an unknown tangent

Given that $\tan A=\dfrac12$, $\tan B=\dfrac13$, and both angles are acute, find $\tan(A+B)$.

$$\begin{aligned} \tan(A+B) &=\frac{\frac12+\frac13}{1-\frac12\cdot\frac13}\\ &=\frac{5/6}{5/6}\\ &=\boxed{1}. \end{aligned}$$

Since $A+B$ is between $0^\circ$ and $180^\circ$ and its tangent is positive, $A+B=45^\circ$. The interval information rules out adding $180^\circ$.

Double angle formulae

Set $B=A=\theta$ in the compound angle formulae. This gives

$$\boxed{\sin2\theta=2\sin\theta\cos\theta},$$ $$\boxed{\cos2\theta=\cos^2\theta-\sin^2\theta},$$

and

$$\boxed{\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}}.$$

For cosine, use $\sin^2\theta+\cos^2\theta=1$ to obtain two alternative forms:

$$\boxed{\cos2\theta=2\cos^2\theta-1=1-2\sin^2\theta.}$$

All three cosine forms are equivalent. The useful form is the one containing the function already present in the problem.

If the problem contains	Prefer
both $\sin\theta$ and $\cos\theta$	$\cos^2\theta-\sin^2\theta$
only $\cos\theta$	$2\cos^2\theta-1$
only $\sin\theta$	$1-2\sin^2\theta$

The square is on the function value

$\cos^2x$ means $(\cos x)^2$. It does not mean $\cos(x^2)$, and $\cos2x$ means $\cos(2x)$, not $2\cos x$.

Worked example 6: derive a double angle result from one ratio

Given that $\sin\theta=\dfrac35$ and $\theta$ is obtuse, find $\sin2\theta$, $\cos2\theta$ and $\tan2\theta$.

Since $\theta$ is obtuse, it lies in the second quadrant, so $\cos\theta<0$. From $\sin^2\theta+\cos^2\theta=1$,

$$\cos\theta=-\sqrt{1-\frac9{25}}=-\frac45.$$

Therefore

$$\sin2\theta=2\sin\theta\cos\theta =2\cdot\frac35\cdot\left(-\frac45\right) =\boxed{-\frac{24}{25}},$$ $$\cos2\theta=1-2\sin^2\theta =1-2\cdot\frac9{25} =\boxed{\frac7{25}},$$

and

$$\tan2\theta=\frac{\sin2\theta}{\cos2\theta} =\boxed{-\frac{24}{7}}.$$

The square root has both signs algebraically. The quadrant information selects the negative cosine.

Self check 3

Given $\cos x=-\dfrac5{13}$ and $\pi<x<\dfrac{3\pi}{2}$, find $\sin2x$ and $\cos2x$.

Answer

The interval places $x$ in the third quadrant, so $\sin x<0$:

$$\sin x=-\sqrt{1-\frac{25}{169}}=-\frac{12}{13}.$$

Hence

$$\sin2x=2\left(-\frac{12}{13}\right)\left(-\frac5{13}\right) =\boxed{\frac{120}{169}},$$ $$\cos2x=2\left(-\frac5{13}\right)^2-1 =\frac{50}{169}-1 =\boxed{-\frac{119}{169}}.$$

Reducing powers

Rearranging the alternative forms of $\cos2x$ gives

$$\boxed{\cos^2x=\frac{1+\cos2x}{2}}, \qquad \boxed{\sin^2x=\frac{1-\cos2x}{2}}.$$

These are often called power reduction formulae. They replace a squared trigonometric function by a first power at twice the angle. They are useful in integration, modelling and identity proofs.

Worked example 7: simplify a squared expression

Express $3\sin^2x-2\cos^2x$ in terms of $\cos2x$.

Substitute both power reduction formulae:

$$\begin{aligned} 3\sin^2x-2\cos^2x &=3\left(\frac{1-\cos2x}{2}\right) -2\left(\frac{1+\cos2x}{2}\right)\\ &=\frac{3-3\cos2x-2-2\cos2x}{2}\\ &=\boxed{\frac{1-5\cos2x}{2}}. \end{aligned}$$

Proving identities

An identity is true for every value in its domain. Begin with one side and transform it into the other. Do not assume the result by performing unrelated steps on both sides.

A reliable strategy is to:

1. expand compound or double angles;
2. rewrite tangent as sine divided by cosine if helpful;
3. use $\sin^2x+\cos^2x=1$;
4. factor or combine fractions;
5. stop as soon as the target appears.

Worked example 8: prove an identity

Prove that

$$\frac{1-\cos2x}{\sin2x}=\tan x$$

where both sides are defined.

Start with the more complicated left hand side. Use $1-\cos2x=2\sin^2x$ and $\sin2x=2\sin x\cos x$:

$$\begin{aligned} \frac{1-\cos2x}{\sin2x} &=\frac{2\sin^2x}{2\sin x\cos x}\\ &=\frac{\sin x}{\cos x}\\ &=\tan x. \end{aligned}$$

Cancellation assumes $\sin x\ne0$, which is already required by the original denominator $\sin2x$ in the relevant cases. The identity is asserted only on its common domain.

Worked example 9: a compound angle proof

Prove that

$$\sin(x+y)\sin(x-y)=\sin^2x-\sin^2y.$$

Expand both factors:

$$\begin{aligned} &\sin(x+y)\sin(x-y)\\ &=(\sin x\cos y+\cos x\sin y) (\sin x\cos y-\cos x\sin y). \end{aligned}$$

This has the form $(a+b)(a-b)=a^2-b^2$, so

$$\begin{aligned} &\sin^2x\cos^2y-\cos^2x\sin^2y\\ &=\sin^2x(1-\sin^2y)-(1-\sin^2x)\sin^2y\\ &=\sin^2x-\sin^2y. \end{aligned}$$

Self check 4

Prove that

$$\frac{\sin2x}{1+\cos2x}=\tan x$$

where both sides are defined.

Answer

Use $\sin2x=2\sin x\cos x$ and $1+\cos2x=2\cos^2x$:

$$\frac{\sin2x}{1+\cos2x} =\frac{2\sin x\cos x}{2\cos^2x} =\frac{\sin x}{\cos x} =\tan x.$$

Solving equations with double angles

First rewrite the equation so that it contains one trigonometric function or one common angle. Then solve over the interval stated in the question. If the equation contains $2x$, transform the interval for $x$ into an interval for $2x$ before listing solutions.

Worked example 10: convert to a quadratic

Solve

$$\cos2x+\cos x=0, \qquad 0\leq x<2\pi.$$

Use $\cos2x=2\cos^2x-1$ because the other term contains $\cos x$:

$$2\cos^2x-1+\cos x=0.$$

Factor:

$$(2\cos x-1)(\cos x+1)=0.$$

Therefore

$$\cos x=\frac12 \quad\text{or}\quad \cos x=-1.$$

On $0\leq x<2\pi$,

$$\boxed{x=\frac{\pi}{3},\ \pi,\ \frac{5\pi}{3}}.$$

Worked example 11: transform the interval

Solve

$$\sin2x=\frac{\sqrt3}{2}, \qquad 0^\circ\leq x\leq180^\circ.$$

Let $\theta=2x$. Doubling every part of the interval gives

$$0^\circ\leq\theta\leq360^\circ.$$

Within this interval,

$$\sin\theta=\frac{\sqrt3}{2} \quad\Longrightarrow\quad \theta=60^\circ,120^\circ.$$

Since $x=\theta/2$,

$$\boxed{x=30^\circ,60^\circ.}$$

Do not lose solutions

Solving $\sin2x=k$ directly from the interval for $x$ commonly misses solutions. The angle $2x$ runs through twice the range of $x$.

Worked example 12: factor before dividing

Solve

$$\sin2x=\sin x, \qquad 0\leq x<2\pi.$$

Use $\sin2x=2\sin x\cos x$:

$$2\sin x\cos x=\sin x.$$

Move everything to one side and factor:

$$\sin x(2\cos x-1)=0.$$

Thus

$$\sin x=0 \quad\text{or}\quad \cos x=\frac12.$$

Therefore

$$\boxed{x=0,\ \frac{\pi}{3},\ \pi,\ \frac{5\pi}{3}}.$$

Dividing the original equation by $\sin x$ would discard the valid solutions $x=0$ and $x=\pi$.

Self check 5

Solve

$$2\sin^2x+3\cos x-3=0, \qquad 0\leq x\leq2\pi.$$

Answer

Replace $\sin^2x$ by $1-\cos^2x$:

$$2(1-\cos^2x)+3\cos x-3=0,$$

so

$$2\cos^2x-3\cos x+1=0.$$

Factor:

$$(2\cos x-1)(\cos x-1)=0.$$

Hence $\cos x=1/2$ or $\cos x=1$. On the stated closed interval,

$$\boxed{x=0,\ \frac{\pi}{3},\ \frac{5\pi}{3},\ 2\pi.}$$

Both endpoints are included.

Choosing an efficient method

Before calculating, identify the structure.

Structure	Useful move
non standard exact angle	split it into standard angles
$\sin x\cos x$	replace with $\tfrac12\sin2x$
$\sin^2x$ or $\cos^2x$	use a power reduction formula
mixture of $\cos2x$ and $\cos x$	use $\cos2x=2\cos^2x-1$
mixture of $\cos2x$ and $\sin x$	use $\cos2x=1-2\sin^2x$
identity with $1\pm\cos2x$	replace by $2\cos^2x$ or $2\sin^2x$
equation with $\sin2x$ and $\sin x$	expand, collect and factor

The formula with the fewest new functions is usually the best choice.

Common misconceptions

• Incorrect: $\cos(A-B)=\cos A\cos B-\sin A\sin B$. Correct: the difference formula has a plus sign.
• Incorrect: $\cos2x=2\cos x$. Correct: $\cos2x=2\cos^2x-1$.
• Incorrect: taking only the positive square root when recovering a missing ratio. Correct: use quadrant or interval information to select the sign.
• Incorrect: dividing an equation by $\sin x$ or $\cos x$ without checking zero. Correct: collect and factor first.
• Incorrect: rounding exact surds midway through a calculation. Correct: retain exact values unless a decimal accuracy is requested.
• Incorrect: solving for $2x$ over the original interval for $x$. Correct: transform the interval as well as the angle.

Mixed self check

1. Find the exact value of $\sin165^\circ$.
2. Given $\tan A=2$ and $\tan B=\dfrac14$, find $\tan(A-B)$.
3. Express $4\cos^2x-3$ in the form $a+b\cos2x$.
4. Solve $\cos2x=\sin x$ for $0\leq x<2\pi$.

Answers

1. Since $165^\circ=120^\circ+45^\circ$,

$$\sin165^\circ =\frac{\sqrt3}{2}\cdot\frac{\sqrt2}{2} +\left(-\frac12\right)\frac{\sqrt2}{2} =\boxed{\frac{\sqrt6-\sqrt2}{4}}.$$

2.

$$\tan(A-B) =\frac{2-1/4}{1+2(1/4)} =\frac{7/4}{3/2} =\boxed{\frac76}.$$

3. Since $\cos^2x=(1+\cos2x)/2$,

$$4\cos^2x-3=2(1+\cos2x)-3 =\boxed{-1+2\cos2x}.$$

4. Use $\cos2x=1-2\sin^2x$:

$$1-2\sin^2x=\sin x.$$

Let $u=\sin x$. Then

$$2u^2+u-1=0, \qquad (2u-1)(u+1)=0.$$

Thus $\sin x=1/2$ or $\sin x=-1$. Therefore

$$\boxed{x=\frac{\pi}{6},\ \frac{5\pi}{6},\ \frac{3\pi}{2}}.$$

Summary

You should now be able to:

• expand $\sin(A\pm B)$, $\cos(A\pm B)$ and $\tan(A\pm B)$ with correct signs;
• calculate exact trigonometric values at compound angles;
• derive and select among the double angle formulae;
• reduce $\sin^2x$ and $\cos^2x$ to expressions involving $\cos2x$;
• prove identities with attention to their domains;
• solve equations without losing solutions or mishandling the interval.

Next, deepen the algebra in trigonometric identities, apply these formulae in trigonometric equations, and learn to combine $a\cos x+b\sin x$ in harmonic form.