Double Angle Formulas MCQ - Practice Questions & Answers

Solve by difficulty

MEDIUM

Which of the following is equal to $\sin\ 2x$ ?

$2\ \sin\ x\ \cos\ x$

$1-2\ \sin^{2}x$

$2\ \cos^{2}x-1$

$\cos^{2}x-\sin^{2}x$

View Solution

If $\cos\ x=\frac{2}{5}$ , the value of $\cos\ 2x$ is

$5$

$-5$

$-\frac{1}{5}$

$\frac{1}{5}$

View Solution

Concepts Covered - 1

Double Angle Formula and Reduction Formula

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(α + β) = sin α cos β + cos α sin β

If we let α = β = θ, then we have

sin(θ + θ) = sin θ cos θ + cos θ sin θ

sin(2θ) = 2 sin θ cos θ

For cosine

cos(α + β) = cos α cos β − sin α sin β,

Letting α = β = θ, we have

cos(θ + θ) = cos θ cos θ − sin θ sin θ

cos(2θ) = cos² θ − sin² θ

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 α = β = θ 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.

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