Double angle formula

Deducing double angle formulae

To deduce the double angle formulae for sine, cosine, and tangent, the compound angle formulae are used. This is done when two angles are the same, i.e., $A = B$.

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

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

This is the double angle formula for sine.

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

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

This is the double angle formula for cosine. This identity can be expressed in several forms:

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

Therefore:

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

Also, from $\tan(A + B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B}$, we have:

$$\begin{aligned} \tan 2A &= \tan(A + A) = \frac{\tan A + \tan A}{1 - \tan A \tan A} \\ \tan 2A &= \frac{2\tan A}{1 - \tan^2 A} \end{aligned}$$

Example 1

Prove that $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$.

Solution:

Use $\sin 3\theta = \sin(2\theta + \theta)$. Then:

$$\begin{aligned} \sin 3\theta &= \sin 2\theta \cos\theta + \cos 2\theta \sin\theta \\ &= (2\sin\theta \cos\theta)\cos\theta + (\cos^2\theta - \sin^2\theta)\sin\theta \\ &= 2\sin\theta \cos^2\theta + \sin\theta \cos^2\theta - \sin^3\theta \\ &= 3\sin\theta \cos^2\theta - \sin^3\theta \end{aligned}$$

But $\cos^2\theta = 1 - \sin^2\theta$.

$$\begin{aligned} \sin 3\theta &= 3\sin\theta(1 - \sin^2\theta) - \sin^3\theta \\ &= 3\sin\theta - 3\sin^3\theta - \sin^3\theta \\ &= 3\sin\theta - 4\sin^3\theta \end{aligned}$$

Hence proved.

Example 2

Find the value of $\theta$ such that $\sin 2\theta - \cos\theta = 0$ for $0° \le \theta \le 360°$.

Solution:

Given $\sin 2\theta - \cos\theta = 0$. Then:

$$\begin{aligned} 2\sin\theta \cos\theta - \cos\theta &= 0 \\ \cos\theta(2\sin\theta - 1) &= 0 \end{aligned}$$

So, $\cos\theta = 0$ or $2\sin\theta - 1 = 0$, which implies $\sin\theta = 1/2$.

In the interval $0° \le \theta \le 360°$:

$\cos\theta = 0$ implies $\theta = 90°$ and $270°$.

$\sin\theta = 1/2$ implies $\theta = 30°$ and $150°$.

Therefore, the values of $\theta$ satisfying the equation $\sin 2\theta - \cos\theta = 0$ for the given interval are $\theta = 30°, 90°, 150°,$ and $270°$.