Compound angle formulae

An angle made by adding or sub­tract­ing other angles, such as $A + B$, $A - B$ or $A + B + C$, is called a com­pound angle. A com­mon mis­take is to think $\sin(A + B) = \sin A + \sin B$; it is not true (try $A = B = 45°$). The cor­rect for­mu­lae are the sub­ject of this les­son. They are the back­bone of the rest of the chap­ter: dou­ble angles, trans­for­ma­tions and almost every iden­tity you will prove come from them.

Group I: sine and cosine of a sum or dif­fer­ence#

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

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

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

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

Notice the pat­tern of signs: the sine for­mu­lae keep the sign of the angle; the cosine for­mu­lae switch it.

Group II: adding and sub­tract­ing the Group I for­mu­lae#

Adding or sub­tract­ing pairs from Group I can­cels half the terms:

$$\sin(A + B) + \sin(A - B) = 2\sin A \cos B$$

$$\sin(A + B) - \sin(A - B) = 2\cos A \sin B$$

$$\cos(A + B) + \cos(A - B) = 2\cos A \cos B$$

$$\cos(A + B) - \cos(A - B) = -2\sin A \sin B$$

$$\cos(A - B) - \cos(A + B) = 2\sin A \sin B$$

These turn a prod­uct into a sum.

Group III: sums into prod­ucts#

In Group II put $A + B = C$ and $A - B = D$. Adding, $2A = C + D$, so $\displaystyle A = \frac{C + D}{2}$; sub­tract­ing, $\displaystyle B = \frac{C - D}{2}$. Then Group II becomes

$$\displaystyle \sin C + \sin D = 2\sin\frac{C + D}{2}\cos\frac{C - D}{2}$$

$$\displaystyle \sin C - \sin D = 2\cos\frac{C + D}{2}\sin\frac{C - D}{2}$$

$$\displaystyle \cos C + \cos D = 2\cos\frac{C + D}{2}\cos\frac{C - D}{2}$$

$$\displaystyle \cos C - \cos D = -2\sin\frac{C + D}{2}\sin\frac{C - D}{2}$$

These turn a sum into a prod­uct. Groups II and III together are called the trans­for­ma­tion for­mu­lae, and a later les­son is devoted to them.

Group IV: tan­gent and cotan­gent#

$$\displaystyle \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}, \qquad \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$$

$$\displaystyle \cot(A + B) = \frac{\cot A \cot B - 1}{\cot B + \cot A}, \qquad \cot(A - B) = \frac{\cot A \cot B + 1}{\cot B - \cot A}$$

Proof of the first. Divide the top and bot­tom of $\displaystyle \frac{\sin(A + B)}{\cos(A + B)}$ by $\cos A \cos B$:

$$\displaystyle \tan(A + B) = \frac{\sin A \cos B + \cos A \sin B}{\cos A \cos B - \sin A \sin B} = \frac{\tan A + \tan B}{1 - \tan A \tan B}$$

For $\cot(A + B)$, take the rec­i­p­ro­cal and write $\displaystyle \tan A = \frac{1}{\cot A}$:

$$\displaystyle \cot(A + B) = \frac{1 - \frac{1}{\cot A \cot B}}{\frac{1}{\cot A} + \frac{1}{\cot B}} = \frac{\cot A \cot B - 1}{\cot B + \cot A}$$

Group V: prod­ucts of a sum and a dif­fer­ence#

$$\sin(A + B)\sin(A - B) = \sin^2 A - \sin^2 B = \cos^2 B - \cos^2 A$$

$$\cos(A + B)\cos(A - B) = \cos^2 A - \sin^2 B = \cos^2 B - \sin^2 A$$

$$\displaystyle \tan(A + B)\tan(A - B) = \frac{\tan^2 A - \tan^2 B}{1 - \tan^2 A \tan^2 B}$$

Proof of the first. Mul­ti­ply out the Group I forms:

$$\begin{aligned} \sin(A + B)\sin(A - B) &= \sin^2 A\cos^2 B - \cos^2 A\sin^2 B \\ &= \sin^2 A(1 - \sin^2 B) - (1 - \sin^2 A)\sin^2 B \\ &= \sin^2 A - \sin^2 B \end{aligned}$$

Group VI: a use­ful tan­gent form#

Divid­ing top and bot­tom by $\cos\theta$, and using $\displaystyle \tan\frac{\pi}{4} = 1$:

$$\displaystyle \frac{\cos\theta + \sin\theta}{\cos\theta - \sin\theta} = \frac{1 + \tan\theta}{1 - \tan\theta} = \tan\left(\frac{\pi}{4} + \theta\right) = \cot\left(\frac{\pi}{4} - \theta\right)$$

$$\displaystyle \frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta} = \frac{1 - \tan\theta}{1 + \tan\theta} = \tan\left(\frac{\pi}{4} - \theta\right) = \cot\left(\frac{\pi}{4} + \theta\right)$$

Group VII: three angles#

Write $\Sigma$ for "the sum of all such terms" and $\Pi$ for "the prod­uct".

$$\sin(A + B + C) = \Sigma \sin A \cos B \cos C - \Pi \sin A$$

Proof. Treat $A + B$ as one angle:

$$\begin{aligned} \sin[(A + B) + C] &= \sin(A + B)\cos C + \cos(A + B)\sin C \\ &= (\sin A\cos B + \cos A\sin B)\cos C + (\cos A\cos B - \sin A\sin B)\sin C \\ &= \sin A\cos B\cos C + \cos A\sin B\cos C + \cos A\cos B\sin C - \sin A\sin B\sin C \end{aligned}$$

which is $\Sigma \sin A \cos B \cos C - \Pi \sin A$. In the same way,

$$\cos(A + B + C) = \Pi\cos A - \Sigma\cos A\sin B\sin C$$

$$\displaystyle \tan(A + B + C) = \frac{\Sigma\tan A - \Pi\tan A}{1 - \Sigma\tan A\tan B} = \frac{\tan A + \tan B + \tan C - \tan A\tan B\tan C}{1 - (\tan A\tan B + \tan B\tan C + \tan C\tan A)}$$

Group VIII: the ratios of $15°$#

Take $A = 60°$, $B = 45°$ in $\sin(A - B)$:

$$\displaystyle \sin 15° = \sin 60°\cos 45° - \cos 60°\sin 45° = \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} - \frac{1}{2} \cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{3} - 1}{2\sqrt{2}}$$

Sim­i­larly,

$$\displaystyle \sin 15° = \frac{\sqrt{3} - 1}{2\sqrt{2}}, \quad \cos 15° = \frac{\sqrt{3} + 1}{2\sqrt{2}}, \quad \tan 15° = 2 - \sqrt{3}, \quad \cot 15° = 2 + \sqrt{3}$$

Since $75° = 90° - 15°$, the ratios of $75°$ are the co-ratios of $15°$: $\displaystyle \sin 75° = \frac{\sqrt{3} + 1}{2\sqrt{2}}$, $\tan 75° = 2 + \sqrt{3}$.

Exam­ple 1#

Find $\cos 105°$.

$$\displaystyle \cos(60° + 45°) = \cos 60°\cos 45° - \sin 60°\sin 45° = \frac{1}{2\sqrt{2}} - \frac{\sqrt{3}}{2\sqrt{2}} = \frac{1 - \sqrt{3}}{2\sqrt{2}}$$

Exam­ple 2#

If $\displaystyle \sin A = \frac{3}{5}$ and $\displaystyle \cos B = \frac{5}{13}$, with $A$ and $B$ acute, find $\sin(A + B)$ and $\cos(A + B)$.

From right tri­an­gles, $\displaystyle \cos A = \frac{4}{5}$ and $\displaystyle \sin B = \frac{12}{13}$.

$$\displaystyle \sin(A + B) = \frac{3}{5} \cdot \frac{5}{13} + \frac{4}{5} \cdot \frac{12}{13} = \frac{15}{65} + \frac{48}{65} = \frac{63}{65}$$

$$\displaystyle \cos(A + B) = \frac{4}{5} \cdot \frac{5}{13} - \frac{3}{5} \cdot \frac{12}{13} = \frac{20}{65} - \frac{36}{65} = -\frac{16}{65}$$

The cosine is neg­a­tive, so $A + B$ is obtuse.

Prac­tice#

1. Find $\cos 75°$.
2. Find $\sin 105°$.
3. Find $\tan 105°$.
4. Find $\sin 40°\cos 20° + \cos 40°\sin 20°$.
5. Find $\cos 70°\cos 10° + \sin 70°\sin 10°$.
6. If $\displaystyle \tan A = \frac{1}{2}$ and $\displaystyle \tan B = \frac{1}{3}$, find $\tan(A + B)$, and hence $A + B$ for acute $A$, $B$.
7. If $\displaystyle \sin A = \frac{4}{5}$ and $\displaystyle \sin B = \frac{5}{13}$, with $A$ and $B$ acute, find $\sin(A - B)$ and $\cos(A - B)$.
8. Find $\sin^2 75° - \sin^2 15°$.
9. Find $\cos^2 45° - \sin^2 15°$.
10. Find $\displaystyle \frac{\cos 15° + \sin 15°}{\cos 15° - \sin 15°}$.
11. Write $\displaystyle \frac{1 - \tan 20°}{1 + \tan 20°}$ as a sin­gle tan­gent.
12. Find $\cot 75°$.

Answers#