Compound Angle Formulas

You know the exact values of sine and cosine for the special angles like $\dfrac{\pi}{6}$, $\dfrac{\pi}{4}$ and $\dfrac{\pi}{3}$. The compound angle formulas let you go further: they give the sine, cosine, or tangent of a sum or difference of two angles, $a + b$ or $a - b$, from the ratios of $a$ and $b$. With them you can find exact values for angles like $\dfrac{\pi}{12}$ and $\dfrac{7\pi}{12}$, simplify expressions, and prove new identities. All angles are in radians.

Key ideas

The formulas

$$\begin{aligned} \sin(a + b) &= \sin a\cos b + \cos a\sin b \\ \sin(a - b) &= \sin a\cos b - \cos a\sin b \\ \cos(a + b) &= \cos a\cos b - \sin a\sin b \\ \cos(a - b) &= \cos a\cos b + \sin a\sin b \\ \tan(a + b) &= \frac{\tan a + \tan b}{1 - \tan a\tan b} \\ \tan(a - b) &= \frac{\tan a - \tan b}{1 + \tan a\tan b} \end{aligned}$$

Things to notice:

• The sine formulas mix the functions ($\sin\cos$ and $\cos\sin$) and keep the sign: $+$ on the left gives $+$ on the right.
• The cosine formulas keep the functions together ($\cos\cos$ and $\sin\sin$) and flip the sign: $+$ on the left gives $-$ on the right.
• The tangent formulas keep the sign on top and flip it on the bottom.

A trig function does not distribute over a sum. For example, with $a = b = \dfrac{\pi}{4}$:

$$\sin\left(\frac{\pi}{4} + \frac{\pi}{4}\right) = \sin\frac{\pi}{2} = 1, \qquad\text{but}\qquad \sin\frac{\pi}{4} + \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2}$$

Where the cosine difference formula comes from

You don’t have to reproduce this argument, but it’s worth following once, because it shows the formulas aren’t magic.

Put two angles $a$ and $b$ in standard position on the unit circle. Their terminal arms meet the circle at $P(\cos a, \sin a)$ and $Q(\cos b, \sin b)$, and the angle between $OP$ and $OQ$ is $a - b$.

Find the length of $PQ$ in two ways: with the distance formula and with the cosine law.

Distance formula:

$$\begin{aligned} PQ^2 &= (\cos a - \cos b)^2 + (\sin a - \sin b)^2 \\ &= \cos^2 a - 2\cos a\cos b + \cos^2 b + \sin^2 a - 2\sin a\sin b + \sin^2 b \\ &= (\sin^2 a + \cos^2 a) + (\sin^2 b + \cos^2 b) - 2\cos a\cos b - 2\sin a\sin b \\ &= 2 - 2(\cos a\cos b + \sin a\sin b) \end{aligned}$$

Cosine law in triangle $OPQ$, where $OP = OQ = 1$:

$$PQ^2 = 1^2 + 1^2 - 2(1)(1)\cos(a - b) = 2 - 2\cos(a - b)$$

Both expressions equal $PQ^2$, so setting them equal and simplifying gives

$$\cos(a - b) = \cos a\cos b + \sin a\sin b$$

The other formulas follow from this one:

• $\cos(a + b)$: write $a + b = a - (-b)$ and use $\cos(-b) = \cos b$ and $\sin(-b) = -\sin b$.
• $\sin(a + b)$: use the cofunction identity, $\sin(a + b) = \cos\left(\dfrac{\pi}{2} - (a + b)\right) = \cos\left(\left(\dfrac{\pi}{2} - a\right) - b\right)$, then the difference formula.
• $\tan(a + b)$: divide $\sin(a + b)$ by $\cos(a + b)$, then divide the top and bottom by $\cos a\cos b$.

Exact values

To find an exact value, write the angle as a sum or difference of two special angles. With a denominator of $12$, the useful pieces are $\dfrac{\pi}{6} = \dfrac{2\pi}{12}$, $\dfrac{\pi}{4} = \dfrac{3\pi}{12}$, $\dfrac{\pi}{3} = \dfrac{4\pi}{12}$, and so on. For example:

$$\frac{\pi}{12} = \frac{4\pi}{12} - \frac{3\pi}{12} = \frac{\pi}{3} - \frac{\pi}{4} \qquad\qquad \frac{7\pi}{12} = \frac{4\pi}{12} + \frac{3\pi}{12} = \frac{\pi}{3} + \frac{\pi}{4}$$

Here are the special values you’ll need (special angles in radians):

$x$	$\dfrac{\pi}{6}$	$\dfrac{\pi}{4}$	$\dfrac{\pi}{3}$
$\sin x$	$\dfrac{1}{2}$	$\dfrac{\sqrt{2}}{2}$	$\dfrac{\sqrt{3}}{2}$
$\cos x$	$\dfrac{\sqrt{3}}{2}$	$\dfrac{\sqrt{2}}{2}$	$\dfrac{1}{2}$
$\tan x$	$\dfrac{1}{\sqrt{3}}$	$1$	$\sqrt{3}$

Using the formulas in reverse

If you see the pattern $\sin a\cos b + \cos a\sin b$, you can collapse it into $\sin(a + b)$. Spotting these patterns makes some ugly-looking expressions very simple.

Worked examples

Example 1: An exact value with sine

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

Solution. Write $\dfrac{\pi}{12} = \dfrac{\pi}{3} - \dfrac{\pi}{4}$ and use the sine difference formula:

$$\begin{aligned} \sin\frac{\pi}{12} &= \sin\left(\frac{\pi}{3} - \frac{\pi}{4}\right) \\ &= \sin\frac{\pi}{3}\cos\frac{\pi}{4} - \cos\frac{\pi}{3}\sin\frac{\pi}{4} \\ &= \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2} - \frac{1}{2}\cdot\frac{\sqrt{2}}{2} \\ &= \frac{\sqrt{6} - \sqrt{2}}{4} \end{aligned}$$

Check: $\dfrac{\sqrt{6} - \sqrt{2}}{4} \approx \dfrac{2.4495 - 1.4142}{4} \approx 0.2588$, and a calculator in radian mode gives $\sin\dfrac{\pi}{12} \approx 0.2588$. ✓

Example 2: An exact value with cosine

Find the exact value of $\cos\dfrac{7\pi}{12}$.

Solution. Write $\dfrac{7\pi}{12} = \dfrac{\pi}{3} + \dfrac{\pi}{4}$. The cosine sum formula has a minus sign:

$$\begin{aligned} \cos\frac{7\pi}{12} &= \cos\frac{\pi}{3}\cos\frac{\pi}{4} - \sin\frac{\pi}{3}\sin\frac{\pi}{4} \\ &= \frac{1}{2}\cdot\frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2} \\ &= \frac{\sqrt{2} - \sqrt{6}}{4} \end{aligned}$$

Check: $\dfrac{7\pi}{12}$ is between $\dfrac{\pi}{2}$ and $\pi$ (quadrant 2), where cosine is negative. Since $\sqrt{6} \gt \sqrt{2}$, the answer is negative. ✓ (It’s about $-0.2588$.)

Example 3: Using a formula in reverse

Simplify, and find the exact value where possible.

(a) $\sin\dfrac{5\pi}{12}\cos\dfrac{\pi}{12} - \cos\dfrac{5\pi}{12}\sin\dfrac{\pi}{12}$

(b) $\cos x\cos 2x - \sin x\sin 2x$

Solution.

(a) This is the pattern $\sin a\cos b - \cos a\sin b = \sin(a - b)$ with $a = \dfrac{5\pi}{12}$ and $b = \dfrac{\pi}{12}$:

$$\sin\left(\frac{5\pi}{12} - \frac{\pi}{12}\right) = \sin\frac{4\pi}{12} = \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$$

(b) This is the pattern $\cos a\cos b - \sin a\sin b = \cos(a + b)$ with $a = x$ and $b = 2x$:

$$\cos x\cos 2x - \sin x\sin 2x = \cos(x + 2x) = \cos 3x$$

Example 4: Using the quadrants

Angle $a$ is in quadrant 1 with $\sin a = \dfrac{3}{5}$, and angle $b$ is in quadrant 2 with $\cos b = -\dfrac{5}{13}$. Find the exact values of $\sin(a + b)$ and $\cos(a + b)$.

Solution. First find the missing ratios with $\sin^2 x + \cos^2 x = 1$, choosing the signs from the quadrants.

$$\cos a = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{\frac{16}{25}} = \frac{4}{5} \qquad \text{(quadrant 1, so positive)}$$ $$\sin b = \sqrt{1 - \left(-\frac{5}{13}\right)^2} = \sqrt{\frac{144}{169}} = \frac{12}{13} \qquad \text{(quadrant 2, sine positive)}$$

Now use the formulas:

$$\begin{aligned} \sin(a + b) &= \sin a\cos b + \cos a\sin b = \frac{3}{5}\left(-\frac{5}{13}\right) + \frac{4}{5}\cdot\frac{12}{13} = \frac{-15 + 48}{65} = \frac{33}{65} \\[4pt] \cos(a + b) &= \cos a\cos b - \sin a\sin b = \frac{4}{5}\left(-\frac{5}{13}\right) - \frac{3}{5}\cdot\frac{12}{13} = \frac{-20 - 36}{65} = -\frac{56}{65} \end{aligned}$$

Check: $\left(\dfrac{33}{65}\right)^2 + \left(-\dfrac{56}{65}\right)^2 = \dfrac{1089 + 3136}{4225} = \dfrac{4225}{4225} = 1$. ✓

Common mistakes

Distributing the function. $\sin(a + b)$ is not $\sin a + \sin b$, and $\cos(a - b)$ is not $\cos a - \cos b$. The numerical check in Key ideas ($1$ versus $\sqrt{2}$) shows why.

Getting the cosine sign backwards. The cosine formulas flip the sign: $\cos(a + b)$ has a minus, and $\cos(a - b)$ has a plus. If you’re unsure, test with $b = 0$ or with special angles you know.

Mixing up the tangent signs. In $\tan(a + b) = \dfrac{\tan a + \tan b}{1 - \tan a\tan b}$, the top keeps the sign and the bottom flips it.

Choosing the wrong sign for a missing ratio. In Example 4, $\sqrt{1 - \cos^2 b}$ gives a size; the quadrant decides the sign. Always state the quadrant and use CAST before you substitute.

Splitting the angle incorrectly. Check your split by adding the fractions back together with a common denominator: $\dfrac{\pi}{3} + \dfrac{\pi}{4} = \dfrac{4\pi + 3\pi}{12} = \dfrac{7\pi}{12}$. ✓

Combining radicals that can’t be combined. $\dfrac{\sqrt{6} - \sqrt{2}}{4}$ does not simplify to $\dfrac{\sqrt{4}}{4}$. You can only add or subtract like radicals.

Practice

1. (Warm-up) Write each angle as a sum or difference of two of the special angles $\dfrac{\pi}{6}$, $\dfrac{\pi}{4}$, $\dfrac{\pi}{3}$, $\dfrac{2\pi}{3}$.

• (a) $\dfrac{5\pi}{12}$
• (b) $\dfrac{11\pi}{12}$
• (c) $-\dfrac{\pi}{12}$

Solution

Use twelfths: $\dfrac{\pi}{6} = \dfrac{2\pi}{12}$, $\dfrac{\pi}{4} = \dfrac{3\pi}{12}$, $\dfrac{\pi}{3} = \dfrac{4\pi}{12}$, $\dfrac{2\pi}{3} = \dfrac{8\pi}{12}$.

(a) $\dfrac{5\pi}{12} = \dfrac{3\pi}{12} + \dfrac{2\pi}{12} = \dfrac{\pi}{4} + \dfrac{\pi}{6}$

(b) $\dfrac{11\pi}{12} = \dfrac{8\pi}{12} + \dfrac{3\pi}{12} = \dfrac{2\pi}{3} + \dfrac{\pi}{4}$

(c) $-\dfrac{\pi}{12} = \dfrac{3\pi}{12} - \dfrac{4\pi}{12} = \dfrac{\pi}{4} - \dfrac{\pi}{3}$

2. (Warm-up) Write as a single trig ratio, then evaluate: $\cos\dfrac{\pi}{5}\cos\dfrac{3\pi}{10} - \sin\dfrac{\pi}{5}\sin\dfrac{3\pi}{10}$.

Solution

This is $\cos(a + b)$ with $a = \dfrac{\pi}{5}$ and $b = \dfrac{3\pi}{10}$:

$$\cos\left(\frac{\pi}{5} + \frac{3\pi}{10}\right) = \cos\left(\frac{2\pi}{10} + \frac{3\pi}{10}\right) = \cos\frac{5\pi}{10} = \cos\frac{\pi}{2} = 0$$

3. (Core) Find the exact value of $\sin\dfrac{5\pi}{12}$.

Solution

$$\begin{aligned} \sin\frac{5\pi}{12} &= \sin\left(\frac{\pi}{4} + \frac{\pi}{6}\right) \\ &= \sin\frac{\pi}{4}\cos\frac{\pi}{6} + \cos\frac{\pi}{4}\sin\frac{\pi}{6} \\ &= \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2}\cdot\frac{1}{2} \\ &= \frac{\sqrt{6} + \sqrt{2}}{4} \end{aligned}$$

Check: $\approx 0.9659$, which matches a calculator. ✓

4. (Core) Find the exact value of $\cos\dfrac{11\pi}{12}$.

Solution

$$\begin{aligned} \cos\frac{11\pi}{12} &= \cos\left(\frac{2\pi}{3} + \frac{\pi}{4}\right) \\ &= \cos\frac{2\pi}{3}\cos\frac{\pi}{4} - \sin\frac{2\pi}{3}\sin\frac{\pi}{4} \\ &= \left(-\frac{1}{2}\right)\frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2} \\ &= \frac{-\sqrt{2} - \sqrt{6}}{4} = -\frac{\sqrt{6} + \sqrt{2}}{4} \end{aligned}$$

Check: $\dfrac{11\pi}{12}$ is in quadrant 2, so cosine should be negative. ✓ (About $-0.9659$.)

5. (Core) Find the exact value of $\tan\dfrac{\pi}{12}$, in simplest form.

Solution

$$\tan\frac{\pi}{12} = \tan\left(\frac{\pi}{3} - \frac{\pi}{4}\right) = \frac{\tan\frac{\pi}{3} - \tan\frac{\pi}{4}}{1 + \tan\frac{\pi}{3}\tan\frac{\pi}{4}} = \frac{\sqrt{3} - 1}{1 + \sqrt{3}}$$

Rationalize by multiplying the top and bottom by $\sqrt{3} - 1$:

$$\frac{(\sqrt{3} - 1)^2}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{3 - 2\sqrt{3} + 1}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$$

Check: $2 - \sqrt{3} \approx 0.2679$, which matches $\tan\dfrac{\pi}{12}$ on a calculator. ✓

6. (Core) Use compound angle formulas to show that each identity is true.

• (a) $\sin\left(x + \dfrac{\pi}{2}\right) = \cos x$
• (b) $\cos(\pi - x) = -\cos x$

Solution

(a) $\sin\left(x + \dfrac{\pi}{2}\right) = \sin x\cos\dfrac{\pi}{2} + \cos x\sin\dfrac{\pi}{2} = \sin x(0) + \cos x(1) = \cos x$

(b) $\cos(\pi - x) = \cos\pi\cos x + \sin\pi\sin x = (-1)\cos x + (0)\sin x = -\cos x$

7. (Core) Angle $a$ is in quadrant 3 with $\cos a = -\dfrac{4}{5}$, and angle $b$ is in quadrant 1 with $\sin b = \dfrac{5}{13}$. Find the exact value of $\cos(a - b)$.

Solution

In quadrant 3, sine is negative: $\sin a = -\sqrt{1 - \dfrac{16}{25}} = -\dfrac{3}{5}$. In quadrant 1, $\cos b = \sqrt{1 - \dfrac{25}{169}} = \dfrac{12}{13}$.

$$\begin{aligned} \cos(a - b) &= \cos a\cos b + \sin a\sin b \\ &= \left(-\frac{4}{5}\right)\frac{12}{13} + \left(-\frac{3}{5}\right)\frac{5}{13} \\ &= \frac{-48 - 15}{65} = -\frac{63}{65} \end{aligned}$$

8. (Challenge) Simplify $\cos\left(x + \dfrac{\pi}{3}\right) + \cos\left(x - \dfrac{\pi}{3}\right)$.

Solution

Expand both:

$$\begin{aligned} &\left(\cos x\cos\frac{\pi}{3} - \sin x\sin\frac{\pi}{3}\right) + \left(\cos x\cos\frac{\pi}{3} + \sin x\sin\frac{\pi}{3}\right) \\ &= 2\cos x\cos\frac{\pi}{3} \\ &= 2\cos x\left(\frac{1}{2}\right) = \cos x \end{aligned}$$

The $\sin x$ terms cancel. Check with $x = 0$: $\cos\dfrac{\pi}{3} + \cos\left(-\dfrac{\pi}{3}\right) = \dfrac{1}{2} + \dfrac{1}{2} = 1 = \cos 0$. ✓

9. (Challenge)

• (a) Show that $\tan\left(x + \dfrac{\pi}{4}\right) = \dfrac{1 + \tan x}{1 - \tan x}$.
• (b) Use part (a) to find the exact value of $\tan\dfrac{5\pi}{12}$.

Solution

(a) Since $\tan\dfrac{\pi}{4} = 1$:

$$\tan\left(x + \frac{\pi}{4}\right) = \frac{\tan x + \tan\frac{\pi}{4}}{1 - \tan x\tan\frac{\pi}{4}} = \frac{\tan x + 1}{1 - \tan x}$$

(b) $\dfrac{5\pi}{12} = \dfrac{\pi}{6} + \dfrac{\pi}{4}$, so use $x = \dfrac{\pi}{6}$, where $\tan\dfrac{\pi}{6} = \dfrac{1}{\sqrt{3}}$:

$$\tan\frac{5\pi}{12} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$$

(multiplying the top and bottom by $\sqrt{3}$). Rationalize:

$$\frac{(\sqrt{3} + 1)^2}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$$

Check: $2 + \sqrt{3} \approx 3.732$, which matches $\tan\dfrac{5\pi}{12}$ on a calculator. ✓