Put $\beta = \alpha$ in the sum formulas and you get formulas for twice an angle. They show up all over calculus and physics, so they are worth knowing on sight.
Key idea
$$\sin 2\theta = 2\sin\theta\cos\theta$$ $$\cos 2\theta = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$$ $$\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}$$
Why it works
$$\sin(\theta + \theta) = \sin\theta\cos\theta + \cos\theta\sin\theta = 2\sin\theta\cos\theta,$$ and $\cos(\theta + \theta) = \cos^2\theta - \sin^2\theta$. Replace $\sin^2\theta$ with $1 - \cos^2\theta$ to get $2\cos^2\theta - 1$, or replace $\cos^2\theta$ with $1 - \sin^2\theta$ to get $1 - 2\sin^2\theta$. Three forms of one fact; pick the one that matches what you know.
Worked examples
Worked example: From one value
$\sin\theta = \frac{3}{5}$ and $\theta$ is in Quadrant II. Find $\sin 2\theta$ and $\cos 2\theta$.
- First $\cos\theta$: $\cos^2\theta = 1 - \frac{9}{25} = \frac{16}{25}$, and cosine is negative in Quadrant II, so $\cos\theta = -\frac{4}{5}$.
- $\sin 2\theta = 2 \cdot \frac{3}{5} \cdot \left(-\frac{4}{5}\right) = -\frac{24}{25}$.
- $\cos 2\theta = 1 - 2\sin^2\theta = 1 - \frac{18}{25} = \frac{7}{25}$.
Answer: $\sin 2\theta = -\frac{24}{25}$, $\cos 2\theta = \frac{7}{25}$
Worked example: Reading a formula backward
Find $2\sin 15^\circ\cos 15^\circ$.
- It has the shape $2\sin\theta\cos\theta$ with $\theta = 15^\circ$.
- So it equals $\sin 30^\circ = \frac{1}{2}$.
Answer: $\frac{1}{2}$
Worked example: Choosing the form
If $\cos\theta = \frac{1}{3}$, find $\cos 2\theta$.
- Only cosine is given, so use $\cos 2\theta = 2\cos^2\theta - 1$.
- $2 \cdot \frac{1}{9} - 1 = -\frac{7}{9}$.
Answer: $-\frac{7}{9}$
Common mistake
Writing $\sin 2\theta = 2\sin\theta$. Doubling the angle does not double the value. Check with $\theta = 90^\circ$: $\sin 180^\circ = 0$, but $2\sin 90^\circ = 2$. (Sine values never exceed $1$, so $2\sin\theta$ cannot be a sine.)