Double angle formulae

In a nutshell

The addition formulae allow you to evaluate trigonometric functions at $$\alpha + \beta$$​, assuming you have their values evaluated at both $$\alpha$$​ and $$\beta$$​. The special cases of these formulae where $$\beta = \alpha$$ are the double angle formulae.

Double angle identities

The addition formulae are:

​$$\begin{aligned}\sin(\alpha + \beta) &\equiv \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)\\\cos(\alpha + \beta) &\equiv \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)\\\tan(\alpha + \beta) &\equiv \dfrac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha)\tan(\beta)}\end{aligned}$$​​

In each of these, let $$\beta = \alpha$$ to derive identities for $$\sin(2\alpha)$$​, $$\cos(2\alpha)$$​ and $$\tan(2\alpha)$$​:

​$$\sin(\alpha + \alpha) \equiv \sin(\alpha)\cos(\alpha) + \cos(\alpha)\sin(\alpha)\\\boxed{\sin(2\alpha) \equiv 2\sin(\alpha)\cos(\alpha)} \quad\quad\quad\quad\quad\quad\quad$$​​

​$$\cos(\alpha + \alpha) \equiv \cos(\alpha)\cos(\alpha) - \sin(\alpha)\sin(\alpha) \quad \quad \quad \quad \quad \quad\\\boxed{\cos(2\alpha) \equiv \cos^2(\alpha) - \sin^2(\alpha)} \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\\boxed{\cos(2\alpha) \equiv 2\cos^2(\alpha) - 1} \quad using \quad \sin^2(\alpha)=1-\cos^2(\alpha) \\\boxed{\cos(2\alpha) \equiv 1 -2 \sin^2(\alpha)} \quad using \quad \cos^2(\alpha)=1-\sin^2(\alpha) \\$$​​

​$$\tan(\alpha + \alpha) \equiv \dfrac{\tan(\alpha) + \tan(\alpha)}{1 - \tan(\alpha)\tan(\alpha)}\\\boxed{\tan(2\alpha) \equiv \dfrac{2\tan(\alpha)}{1 - \tan^2(\alpha)}} \quad\quad\quad\quad$$​​

Example 1

Compute the exact value of $$\left(\sin\left(\dfrac{\pi}{8}\right)+ \cos\left(\dfrac{\pi}{8}\right)\right)^2$$.

Expand the brackets, and use the identity $$\cos^2(\alpha) + \sin^2(\alpha) \equiv 1$$ and the double angle identity for $$\sin(2\alpha)$$​:

$$\begin{aligned}\left(\sin\left(\dfrac{\pi}{8}\right) + \cos\left(\dfrac{\pi}{8}\right)\right)^2 &= \sin^2\left(\dfrac{\pi}{8}\right) + \cos^2\left(\dfrac{\pi}{8}\right) + 2\sin\left(\dfrac{\pi}{8}\right)\cos\left(\dfrac{\pi}{8}\right)\\&= 1 + \sin\left(2 \times\dfrac{\pi}{8}\right)\\&= 1 + \sin\left(\dfrac{\pi}{4}\right)\\&= 1 + \dfrac{\sqrt{2}}{2}\\&= \dfrac{2 + \sqrt{2}}{2}\end{aligned}$$​​

Therefore, the exact value of $$\left(\sin\left(\dfrac{\pi}{8}\right) + \cos\left(\dfrac{\pi}{8}\right)\right)^2$$ is $$\underline{\dfrac{2 + \sqrt{2}}{2}}$$.

Example 2

Show that $$\sin(4\alpha) \equiv 4\sin(\alpha)\cos^3(\alpha) - 4\sin^3(\alpha)\cos(\alpha)$$.

​

You have that $$4\alpha = 2 \times 2\alpha$$, so by the double angle formula for $$\sin$$​:

$$\sin(4\alpha) \equiv 2\sin(2\alpha)\cos(2\alpha)$$​​

Apply both the double angle formulae for $$\sin$$​ and $$\cos$$​:

$$\begin{aligned}\sin(4\alpha) &\equiv 2\sin(2\alpha)\cos(2\alpha)\\&\equiv 2(2\sin(\alpha)\cos(\alpha))(\cos^2(\alpha) - \sin^2(\alpha))\\&\equiv 4\sin(\alpha)\cos^3(\alpha) - 4\sin^3(\alpha)\cos(\alpha)\end{aligned}$$​​

Therefore, $$\underline{\sin(4\alpha) \equiv 4\sin(\alpha)\cos^3(\alpha) - 4\sin^3(\alpha)\cos(\alpha)}$$.