Question

The value of $\frac{1}{8}(3-4\cos 2\mathrm{\theta }+\cos 4\mathrm{\theta })$ is

• A. $\cos 4\mathrm{\theta }$
• B. $\sin 4\mathrm{\theta }$
• C. ${\sin }^4\mathrm{\theta }$
• D. ${\cos }^4\mathrm{\theta }$
• E. ${\sin }^4\left(\raisebox{1ex}{$\mathrm{\theta }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$

Answer 1

The correct option is C

${\sin }^4\mathrm{\theta }$

The explanation for the correct answer.

Given: $\frac{1}{8}(3-4\cos 2\mathrm{\theta }+\cos 4\mathrm{\theta })$

Formula $\cos 2\mathrm{A}=2{\cos }^2\mathrm{A}-1$

$\begin{array}{rcl}\frac{1}{8}(3-4\cos 2\mathrm{\theta }+\cos 4\mathrm{\theta })& =& \frac{1}{8}\left(3-4\cos 2\mathrm{\theta }+2{\cos }^{2}2\mathrm{\theta }-1\right)\\ & =& \frac{1}{8}\left(2{\cos }^{2}2\mathrm{\theta }-2·2\cos 2\theta +2\right)\\ & =& \frac{2}{8}\left({\cos }^{2}2\mathrm{\theta }-2\cos 2\theta +1\right)\\ & =& \frac{1}{4}{\left(\cos 2\mathrm{\theta }-1\right)}^2\left[\because {\mathrm{a}}^2-2\mathrm{ab}+{\mathrm{b}}^2={\left(\mathrm{a}-\mathrm{b}\right)}^2\right]\\ & =& \frac{1}{4}{\left(1-2{\sin }^2\mathrm{\theta }-1\right)}^2\left[\because \cos 2\mathrm{\theta }=1-2{\sin }^2\mathrm{\theta }\right]\\ & =& \frac{1}{4}\left(4{\sin }^4\mathrm{\theta }\right)\\ & =& {\sin }^4\mathrm{\theta }\end{array}$

Hence $o\mathrm{ption}\left(\mathrm{C}\right)$ is the correct answer.