Double Angle Formulas - Practice Questions & MCQ

Double Angle Formula and Reduction Formula

$\begin{array}{rl}\sin (2\theta )& =2\sin \theta \cos \theta \\ & =\frac{2\tan \theta }{1+{\tan }^2\theta }\\ \cos (2\theta )& ={\cos }^2\theta -{\sin }^2\theta \\ & =1-2{\sin }^2\theta \\ & =2{\cos }^2\theta -1\\ & =\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\\ \tan (2\theta )& =\frac{2\tan \theta }{1-{\tan }^2\theta }\end{array}$

Proof:

The double-angle formulas are a special case of the sum formulas, where α = β.

For sine

$\sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta$
If we let $\alpha =\beta =\theta$, then we have

$\begin{array}{rl}& \sin (\theta +\theta )=\sin \theta \cos \theta +\cos \theta \sin \theta \\ & \sin (2\theta )=2\sin \theta \cos \theta \end{array}$
For cosine

$\cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta$
Letting $\alpha =\beta =\theta$, we have

$\begin{array}{rl}& \cos (\theta +\theta )=\cos \theta \cos \theta -\sin \theta \sin \theta \\ & \cos (2\theta )={\cos }^2\theta -{\sin }^2\theta \end{array}$

We can write this formula in different forms as per the requirement of the question,

$\begin{array}{rl}\cos (2\theta )& ={\cos }^2\theta -{\sin }^2\theta \\ & =(1-{\sin }^2\theta )-{\sin }^2\theta \\ & =1-2{\sin }^2\theta \end{array}$
The second variation is:

$\begin{array}{rl}\cos (2\theta )& ={\cos }^2\theta -{\sin }^2\theta \\ & ={\cos }^2\theta -(1-{\cos }^2\theta )\\ & =2{\cos }^2\theta -1\end{array}$

For tan
Replacing $\alpha =\beta =\theta$ in the sum formula gives

$\begin{array}{rl}\tan (\alpha +\beta )& =\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }\\ \tan (\theta +\theta )& =\frac{\tan \theta +\tan \theta }{1-\tan \theta \tan \theta }\\ \tan (2\theta )& =\frac{2\tan \theta }{1-{\tan }^2\theta }\end{array}$
Reduction Formula
The double-angle formulas can be used to derive the reduction formulas, which are formulas we can use to reduce the power of a given expression involving even powers of sine or cosine.

$\begin{array}{rl}& {\sin }^2\theta =\frac{1-\cos (2\theta )}{2}\\ & {\cos }^2\theta =\frac{1+\cos (2\theta )}{2}\\ & {\tan }^2\theta =\frac{1-\cos (2\theta )}{1+\cos (2\theta )}\end{array}$