What are double angle identities?

Double angle identities appear in both the AA SL and AA HL courses, but do not appear in either the AI SL or AI HL courses. There are double angle identities for sine and cosine, which appear in both the AA SL and AA HL courses, and a double angle identity for tangent, which appears only in the AA HL course.

There is one double angle identity for the sine ratio

• $\sin 2\theta = 2 \sin \theta \cos \theta$

There are three double angle identities for the cosine ratio

• $\cos 2\theta = \cos ^2 \theta - \sin ^2 \theta$

• $\cos 2\theta = 2 \cos ^2 \theta - 1$

• $\cos 2\theta = 1 - 2 \sin ^2 \theta$

There is one double angle identity for the tangent ratio

• $\tan 2\theta = \dfrac{2 \tan \theta}{1 - \tan ^2 \theta}$

The double angle identities give us a way of expressing a trigonometric ratio in another form that may make it easier to solve an equation, or help us to find the exact value of a new angle, particularly in paper 1 where we cannot use a G.D.C. to help us.

For example, suppose we are asked to solve the equation $\sin \theta = \dfrac{1}{4 \cos \theta}$ for $0 \leq \theta \leq 2\pi$.

We could rearrange this by multiplying both sides by $2 \cos \theta$, giving us

$$\begin{align*} \sin \theta \times 2 \cos \theta &= \dfrac{1}{4 \cos \theta} \times 2 \cos \theta \\[8pt] 2 \sin \theta \cos \theta &= \dfrac{1}{2} \end{align*}$$

Now on the left hand side we can apply the double angle identity for sine so that we have

$$\begin{align*} \sin 2\theta &= \dfrac{1}{2} \end{align*}$$

From here we can proceed as though solving a “normal” trigonometric equation, finding that the two solutions are $\theta = \dfrac{\pi}{12}$ and $\theta = \dfrac{5\pi}{12}$.

As a second example, suppose we are asked to find the exact value of $\tan \left(\dfrac{\pi}{12}\right)$.

From ratio triangles, we know that $\tan \left(\dfrac{\pi}{6}\right)=\dfrac{1}{\sqrt 3}$ and $\dfrac{\pi}{6}$ is double the angle $\dfrac{\pi}{12}$ so we could use the double angle identity for tangent to help us. We can set up our equation as follows:

$$\begin{align*} \tan 2\theta &= \frac{2 \tan \theta}{1 - \tan ^2 \theta} \\[12pt] \tan \left(\frac{\pi}{6}\right) &= \frac{2 \tan \frac{\pi}{12}}{1 - \tan ^2 \frac{\pi}{12}} \\[12pt] \frac{1}{\sqrt 3} &= \frac{2 \tan \frac{\pi}{12}}{1 - \tan ^2 \frac{\pi}{12}} \end{align*}$$

Now we could rearrange this to make a quadratic equation.

$$\begin{align*} \tan ^2 \left(\frac{\pi}{12}\right) + 2\sqrt{3} \tan \left(\frac{\pi}{12}\right) - 1 = 0 \end{align*}$$

We could then complete the square, or use the quadratic formula to find that $\tan\left(\dfrac{\pi}{12}\right) = \pm 2 - \sqrt {3}$.

Then we can reason that $\dfrac{\pi}{12}$ is in quadrant 1 where the tangent function is positive, hence $\tan\left(\dfrac{\pi}{12}\right)$ must equal $2 - \sqrt{3}$.

If faced with a trigonometric equation or expression in an exam that doesn’t appear to be solvable or easy to simplify in its current form,, it is useful to investigate whether a substitution using a double angle identity makes the equation more easily solvable, or the expression more easy to simplify.