Double Angle Formulae

Video lesson · from 40:104 worked examples

One lesson video covers all of Trigonometry, so it opens at 40:10 for this topic — not from the beginning.

We can derive the Double Angle Formulae by replacing $B$ with $A$ in our addition formulae.

\sin 2A = 2\sin A\cos A

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

The three versions of the $\cos 2A$ formula are derived using the National 5 identity $\sin^2 A + \cos^2 A = 1$ .

$\sin 2x$ is not $2\sin x$ : it is $2\sin x\cos x$ . The same applies to cosine — doubling the angle is not doubling the function.
Three versions of $\cos 2x$ : pick the one that leaves the equation in a single trig function. Choosing badly turns an easy question into a hard one.
The formulae work both ways: you may need to replace $2\sin x\cos x$ with $\sin 2x$ to simplify an expression.
They apply to any doubled angle: $\sin 4x = 2\sin 2x\cos 2x$ , not just $2x$ .

Example 1

a) Write down the formula for $\sin 2x$ .
b) Write down the formula for $\sin 6x$ .

a) $\sin 2x = 2\sin x \cos x$

b) We can treat $6x$ as double $3x$ . Let $A = 3x$ :

\sin 6x = 2\sin 3x \cos 3x

Example 2

Simplify $\cos^2\left(\frac{\pi}{6}\right) - \sin^2\left(\frac{\pi}{6}\right)$ .

This matches the structure of $\cos^2 A - \sin^2 A = \cos 2A$ , where $A = \frac{\pi}{6}$ .

\begin{aligned} \cos^2\left(\frac{\pi}{6}\right) - \sin^2\left(\frac{\pi}{6}\right) &= \cos\left(2\left(\frac{\pi}{6}\right)\right) \\ &= \cos\left(\frac{2\pi}{6}\right) \\ &= \cos\left(\frac{\pi}{3}\right) \\ &= \frac{1}{2} \end{aligned}

Example 3

If $\tan p = \frac{4}{3}$ for acute angle $p$ , find the exact values of $\sin 2p$ and $\cos 2p$ .

First, set up a right-angled triangle with Opposite = 4, Adjacent = 3.

Hypotenuse = $\sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ .

Therefore, $\sin p = \frac{4}{5}$ and $\cos p = \frac{3}{5}$ .

\begin{aligned} \sin 2p &= 2\sin p \cos p \\ &= 2\left(\frac{4}{5}\right)\left(\frac{3}{5}\right) \\ &= \frac{24}{25} \end{aligned}

You can use any of the three $\cos 2A$ formulae. Let's use $\cos^2 A - \sin^2 A$ :

\begin{aligned} \cos 2p &= \cos^2 p - \sin^2 p \\ &= \left(\frac{3}{5}\right)^2 - \left(\frac{4}{5}\right)^2 \\ &= \frac{9}{25} - \frac{16}{25} \\ &= -\frac{7}{25} \end{aligned}

Example 4

Given that $\cos 2x = \frac{5}{13}$ , find the exact values of $\sin x$ and $\cos x$ (assume $x$ is acute).

Use the double angle formulae that only contain one term:

To find $\cos x$ :

\begin{aligned} 2\cos^2 x - 1 &= \frac{5}{13} \\ 2\cos^2 x &= \frac{18}{13} \\ \cos^2 x &= \frac{9}{13} \\ \cos x &= \frac{3}{\sqrt{13}} \end{aligned}

To find $\sin x$ :

\begin{aligned} 1 - 2\sin^2 x &= \frac{5}{13} \\ 2\sin^2 x &= \frac{8}{13} \\ \sin^2 x &= \frac{4}{13} \\ \sin x &= \frac{2}{\sqrt{13}} \end{aligned}