When should I use which form of $\cos(2 heta)$?

Use $2\cos^2 heta - 1$ if the rest of the equation contains $\cos heta$. Use $1 - 2\sin^2 heta$ if it contains $\sin heta$. Use $\cos^2 heta - \sin^2 heta$ if it can be factored with a corresponding $\cos heta \pm \sin heta$ term.

Can double angle formulas be applied to $4 heta$?

Yes. Treat $4 heta$ as $2(2 heta)$. For example, $\sin(4 heta) = 2\sin(2 heta)\cos(2 heta)$.

Do these formulas apply to fractional angles?

Yes. The formulas represent a ratio. Substituting $\alpha = heta/2$ yields expressions for $\sin heta$ and $\cos heta$ in terms of $\alpha$, which directly constructs the standard half-angle formulas.

How to use the double angle formulas

The double angle formulas express trigonometric functions of $2 heta$ in terms of functions of $heta$ . They apply when a trigonometric expression or equation contains both a base angle and its double, allowing you to standardize the arguments before algebraic manipulation.

The setup

Memorize the core identities: $\sin(2 heta) = 2\sin heta\cos heta$ $\cos(2 heta) = \cos^2 heta - \sin^2 heta = 2\cos^2 heta - 1 = 1 - 2\sin^2 heta$ $an(2 heta) = \frac{2 an heta}{1 - an^2 heta}$

The steps

Identify mismatched arguments where one angle is exactly twice another (e.g., $heta$ and $2 heta$ ).
Select the identity that matches the other trigonometric functions in the equation. For $\cos(2 heta)$ , choose the form that results in a single function type.
Substitute the chosen identity to eliminate the $2 heta$ argument.
Use standard algebraic techniques (factoring, quadratic formula) to solve the resulting single-angle equation.

Checking the result

Verify all roots by substituting them back into the original equation. Ensure that angles fall within the specified domain (typically $0 \le heta < 2\pi$ ). Evaluate exact values algebraically.

Common errors

Dividing both sides of an equation by a trigonometric function (like $\cos heta$ ) instead of factoring it out, which discards valid roots where that function equals zero. Choosing the wrong form of $\cos(2 heta)$ , which introduces unnecessary variables and prevents factoring.

Worked example

Solve $\sin(2 heta) = \cos heta$ for $0 \le heta < 2\pi$ .

$\sin(2 heta) = \cos heta$ $2\sin heta\cos heta = \cos heta$ $2\sin heta\cos heta - \cos heta = 0$ $\cos heta(2\sin heta - 1) = 0$

Case 1: $\cos heta = 0$ $heta = \frac{\pi}{2}, \frac{3\pi}{2}$

Case 2: $2\sin heta - 1 = 0$ $\sin heta = \frac{1}{2}$ $heta = \frac{\pi}{6}, \frac{5\pi}{6}$

Solution set: $\{ \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{3\pi}{2} \}$

FAQ

Run your own problem

Answer form

References: OpenStax Precalculus, Chapter 7: Trigonometric Identities and Equations · Khan Academy, Trigonometry: Double-angle identities · Stewart Calculus, Appendix D: Trigonometry