Question

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

• A. $\cos 4\theta$
• B. $\sin 4\theta$
• C. ${{\sin }^{4}}\theta$
• D. ${{\cos }^{4}}\theta$

Answer 1

Answer: (C)

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

Answer 2

Given :
$\frac{1}{8}(3-4\cos 2\theta +\cos 4\theta )$
$\Rightarrow$ $\frac{1}{8}\\{3-3\cos 2\theta +(\cos 4\theta -\cos 2\theta )\\}$
=\frac{1}{8}\left\\{ 3(1-\cos 2\theta )+2\sin \left( \frac{2\theta -4\theta }{2} \right).\sin \left( \frac{6\theta }{2} \right) \right\\}
$=\frac{1}{8}\\{6{{\sin }^{2}}\theta -2\sin \theta .\sin 3\theta \\}$
$=\frac{1}{8}\\{6{{\sin }^{2}}\theta -2\sin \theta (3\sin \theta -4{{\sin }^{3}}\theta )\\}$
$=\frac{1}{8}\\{6{{\sin }^{2}}\theta -6{{\sin }^{2}}\theta +8{{\sin }^{4}}\theta \\}={{\sin }^{4}}\theta$
So, the correct option is (C) : ${{\sin }^{4}}\theta$