# Compound Angle Identities

$\sin(\alpha + \beta)$ $\cos(\alpha + \beta)$ $\tan(\alpha + \beta)$

## Geometric Derivation

$\sin(\alpha + \beta) =$ $\sin{\alpha} \cos{\beta} + \cos{\alpha} \sin{\beta}$
$\cos(\alpha + \beta) =$ $\cos{\alpha} \cos{\beta} - \sin{\alpha} \sin{\beta}$

### Exercises

By setting $\theta := \alpha = \beta$, deduce formulas for $\sin{2\theta}$ and $\cos{2\theta}$.
Answer \begin{align*} \sin{2\theta} &= 2\sin{\theta}\cos{\theta} \\ \cos{2\theta} &= \cos^2{\theta} - \sin^2{\theta} \\ &= 1 - 2\sin^2{\theta} \\ &= 2\cos^2{\theta} - 1 \end{align*}

Find $\sin{15\unicode{xb0}}$.
Hint $15\unicode{xb0} = 60\unicode{xb0} + \left(-45\unicode{xb0}\right)$.
Answer \begin{align*} \sin{15\unicode{xb0}} &= \sin(60\unicode{xb0} - 45\unicode{xb0}) \\ &= \sin{60\unicode{xb0}}\cos{-45\unicode{xb0}} + \cos{60\unicode{xb0}}\sin{-45\unicode{xb0}} \\ &= \sin{60\unicode{xb0}}\cos{45\unicode{xb0}} - \cos{60\unicode{xb0}}\sin{45\unicode{xb0}} \\ &= \frac{\sqrt{3}}{2} \frac{\sqrt{2}}{2} - \frac{1}{2} \frac{\sqrt{2}}{2} \\ &= \frac{\sqrt{6} - \sqrt{2}}{4} \\ \end{align*}

## Complex Number Derivation

\begin{align*} e^{i\left(\alpha + \beta\right)} &= e^{i\alpha} e^{i\beta} \\ \cos(\alpha + \beta) + i \sin(\alpha + \beta) &= \left(\cos{\alpha} + i \sin{\alpha}\right) \left(\cos{\beta} + i \sin{\beta}\right) \\ &= \left(\cos{\alpha}\cos{\beta} - \sin{\alpha}\sin{\beta}\right) + i\left(\sin{\alpha}\cos{\beta} + \cos{\alpha}\sin{\beta}\right) \end{align*}

### Exercises

By manipulating the equation $e^{i\left(3\theta\right)} = \left(e^{i\theta}\right)^3$, derive formulas for $\sin{3\theta}$ and $\cos{3\theta}$.
Hint Use binomial expansion, then group the real and imaginary terms.
Answer \begin{align*} e^{i\left(3\theta\right)} &= \left(e^{i\theta}\right)^3 \\ \cos{3\theta} + i \sin{3\theta} &= \left(\cos{\theta} + i \sin{\theta}\right)^3 \\ &= \left(\cos^3{\theta} - 3 \cos{\theta} \sin^2{\theta}\right) + i \left(3\cos^2{\theta} \sin{\theta} - \sin^3{\theta}\right) \\ \end{align*} Therefore, \begin{align*} \cos{3\theta} &= \cos^3{\theta} - 3 \cos{\theta} \sin^2{\theta} \\ &= 4\cos^3{\theta} - 3\cos{\theta} \\ \sin{3\theta} &= 3\cos^2{\theta} \sin{\theta} - \sin^3{\theta} \\ &= 3\sin{\theta} - 4\sin^3{\theta} \end{align*}

Find all roots of the equation $8x^3 - 6x - 1 = 0$.
Hint 1 We know that $\cos{3\theta} = 4\cos^3{\theta} - 3\cos{\theta}$ from the last exercise.
Hint 2 Rewrite as $4x^3 - 3x = \frac{1}{2}$, then substitute $x = \cos{\theta}$.
Answer Let $x = \cos{\theta}$. \begin{align*} \frac{1}{2} &= 4x^3 - 3x \\ &= 4\cos^3{\theta} - 3\cos{\theta} \\ &= \cos{3\theta} \end{align*} Since $\cos{\frac{\pi}{3}} = \frac{1}{2}$, we have $3\theta = \frac{\pi}{3} + 2\pi k$. Three possible values of $\theta$ are: \begin{align*} \theta = \frac{1}{3}\left(\frac{\pi}{3}\right) &\implies x = \cos{\frac{\pi}{9}} \\ \theta = \frac{1}{3}\left(\frac{\pi}{3} + 2\pi\right) &\implies x = \cos{\frac{7\pi}{9}} \\ \theta = \frac{1}{3}\left(\frac{\pi}{3} + 4\pi\right) &\implies x = \cos{\frac{13\pi}{9}} \end{align*} You can solve all cubic equations with this method (after applying a simple horizontal translation and/or stretch).

## Matrix Derivation

$R_\theta := \begin{bmatrix} \cos{\theta} & -\sin{\theta} \\ \sin{\theta} & \cos{\theta} \end{bmatrix}$
\begin{align*} R_{\alpha + \beta} &= R_{\alpha} R_{\beta} \\ \begin{bmatrix} \cos(\alpha + \beta) & -\sin(\alpha + \beta) \\ \sin(\alpha + \beta) & \cos(\alpha + \beta) \end{bmatrix} &= \begin{bmatrix} \cos{\alpha} & -\sin{\alpha} \\ \sin{\alpha} & \cos{\alpha} \end{bmatrix} \begin{bmatrix} \cos{\beta} & -\sin{\beta} \\ \sin{\beta} & \cos{\beta} \end{bmatrix} \\ &= \begin{bmatrix} \cos{\alpha}\cos{\beta} - \sin{\alpha}\sin{\beta} & -(\sin{\alpha}\cos{\beta} + \cos{\alpha}\sin{\beta}) \\ \sin{\alpha}\cos{\beta} + \cos{\alpha}\sin{\beta} & \cos{\alpha}\cos{\beta} - \sin{\alpha}\sin{\beta} \end{bmatrix} \end{align*}

## Tangent Compound Angle Identity

We can derive the tangent angle sum identity as follows: \begin{align*} \tan(\alpha + \beta) &= \frac{\sin(\alpha + \beta)}{\cos(\alpha + \beta)} \\ &= \frac{\sin{\alpha}\cos{\beta} + \cos{\alpha}\sin{\beta}}{\cos{\alpha}\cos{\beta} - \sin{\alpha}\sin{\beta}} \\ &= \frac{\sin{\alpha}\cos{\beta} + \cos{\alpha}\sin{\beta}}{\cos{\alpha}\cos{\beta} - \sin{\alpha}\sin{\beta}} \div \frac{\cos{\alpha}\cos{\beta}}{\cos{\alpha}\cos{\beta}} \\ &= \frac{\tan{\alpha} + \tan{\beta}}{1 - \tan{\alpha}\tan{\beta}} \end{align*}

### Exercises

By setting $\theta := \alpha = \beta$, deduce a formula for $\tan{2\theta}$.
Answer $\tan{2\theta} = \frac{2\tan{\theta}}{1 - \tan^2{\theta}}$

Given that $\tan{\theta} = \frac{1}{5}$, find $\tan{2\theta}$. What do you notice about $\tan{2\theta}$?
Hint Use the previous exercise.
Look at the numerator and denominator of $\tan{2\theta}$. Where have you seen these numbers before?
Answer \begin{align*} \tan{2\theta} &= \frac{2 \times \frac{1}{5}}{1 - \left(\frac{1}{5}\right)^2} \\ &= \frac{\frac{2}{5}}{\frac{24}{25}} \\ &= \frac{5}{12} \\ \end{align*} Note that 5 and 12 are the legs of a Pythagorean triple.
In general, if $\tan{\theta} \in \mathbb{Q}$, then $\tan{2\theta} = \frac{a}{b}$ where $a$ and $b$ are the legs of a Pythagorean triple.
You can generate all possible Pythagorean triples in this way.

## Half-Angle Identities

We can rewrite the cosine double-angle identity with $\frac{\theta}{2}$ in place of $\theta$. \begin{align*} \cos{\theta} &= 2\cos^2{\frac{\theta}{2}} - 1 \\ &= 1 - 2 \sin^2{\frac{\theta}{2}} \end{align*} From this, we can rearrange to obtain $\cos{\frac{\theta}{2}} = \pm \sqrt{\frac{1 + \cos{\theta}}{2}}$ $\sin{\frac{\theta}{2}} = \pm \sqrt{\frac{1 - \cos{\theta}}{2}}$ The $\pm$ signs are necessary since $\cos(\theta) = \cos(\theta + 2 \pi)$, but $\cos{\frac{\theta}{2}} \neq \cos{\frac{\theta + 2\pi}{2}}$ (and likewise for $\sin$).

### Exercise

From the above half-angle identities, derive a formula for $\tan{\frac{\theta}{2}}$ in terms of $\cos{\theta}$.
Hint $\tan{x} = \frac{\sin{x}}{\cos{x}}$
Answer \begin{align*} \tan{\frac{\theta}{2}} &= \frac{\sin{\frac{\theta}{2}}}{\cos{\frac{\theta}{2}}} \\ &= \pm \sqrt{\frac{1 - \cos{\theta}}{1 + \cos{\theta}}} \end{align*}

## Practice Problems

Problems are listed in order of increasing difficulty.
• Review Set 13B, 3
• Review Set 13A, 9
• Exercise 13E, 5
• Review Set 13B, 12
• Review Set 13A, 11
• Review Set 13A, 8
• Exercise 13E, 19
• Review Set 13C, 12