Double Angle Formulas - Practice Questions & MCQ

Double Angle Formula and Reduction Formula

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

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

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

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

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

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

$
\begin{aligned}
\cos (2 \theta) & =\cos ^2 \theta-\sin ^2 \theta \\
& =\cos ^2 \theta-\left(1-\cos ^2 \theta\right) \\
& =2 \cos ^2 \theta-1
\end{aligned}
$

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

$
\begin{aligned}
\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{aligned}
$
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{aligned}
& \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{aligned}
$