Question

${\cos }^4\theta -{\sin }^4\theta$ is equal to

• A. $1-2{\sin }^2\left(\frac{\theta }{2}\right)$
• B. $2{\cos }^2\theta -1$
• C. $1+2{\sin }^2\left(\frac{\theta }{2}\right)$
• D. $1+2{\cos }^2\theta$

Answer 1

The correct option is B

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

Explanation for the correct option:

Given, ${\cos }^4\theta –{\sin }^4\theta$

$={\left({\cos }^2\theta \right)}^2–{\left({\sin }^2\theta \right)}^2$

$=({\cos }^2\theta +{\sin }^2\theta )({\cos }^2\theta –{\sin }^2\theta )$ ; $\left[\mathbf{\because }{\mathit{a}}^{\mathbf{2}}\mathbf{}\mathbf{–}\mathbf{}{\mathit{b}}^{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{(}\mathit{a}\mathbf{}\mathbf{+}\mathbf{}\mathit{b}\mathbf{)}\mathbf{(}\mathit{a}\mathbf{}\mathbf{–}\mathbf{}\mathit{b}\mathbf{)}\right]$

$=1\times ({\cos }^2\theta -{\sin }^2\theta )$ ; $\left[\mathbf{\because }\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{\theta }\mathbf{+}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\right]$

$=\cos 2\theta$ ; $\left[\mathbf{\because }\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{\theta }\mathbf{-}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{2}\mathit{\theta }\right]$

$=\mathbf{2}\mathbf{}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{\theta }\mathbf{–}\mathbf{1}$ ; $\left[\mathbf{\because }\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{2}\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{\theta }\mathbf{-}\mathbf{1}\right]$

Hence, Option ‘B’ is Correct.