Double angles: sin 2A, cos 2A and tan 2A

Double-angle questions mix an angle and double that angle: $x$ and $2x$, or $\theta$ and $2\theta$.

Your job is to make every term a function of the same angle. Then it's an ordinary trig equation.

Putting $B = A$ in $\cos(A + B)$ gives the first form:

$\cos 2A = \cos^2 A - \sin^2 A$

The other two swap one square using $\sin^2 A + \cos^2 A = 1$:

• Replace $\cos^2 A$ with $1 - \sin^2 A$: $\;\cos 2A = 1 - 2\sin^2 A$.
• Replace $\sin^2 A$ with $1 - \cos^2 A$: $\;\cos 2A = 2\cos^2 A - 1$.

Each form is best for a different job:

Pick the form that leaves only one trig function in the equation.

1. Look at the other terms. Do they contain $\sin A$ or $\cos A$?
2. Replace the $2A$ term with the form that matches. Now every term uses $A$.
3. Solve the result, which is often a quadratic in $\sin A$ or $\cos A$.
4. Find every angle $A$ in the given interval, and check one in the original equation.

Note: once you've rewritten everything in terms of $A$, you solve over the interval you were given. Doubling the interval only applies when you solve for $2A$ directly. For example, in $\sin 2x = 0.5$ for $0° \le x \le 360°$, the angle $2x$ runs from $0°$ to $720°$, so you find all values of $2x$ in that range and then halve them.