Question

The value of $\frac{1}{8}(3-4cos2\theta +cos4\theta )$ is

• A. $\cos 4\theta$
• B. $\sin 4\theta$
• C. ${\sin }^4\theta$
• D. ${\cos }^4\theta$
• E. ${\sin }^4(\theta /2)$

Answer 1

Answer: (C)

$\frac{1}{8}(3-4\cos 2\theta +\cos 4\theta )$
$\Rightarrow$ $\frac{1}{8}\{3-3\cos 2\theta +(\cos 4\theta -\cos 2\theta )\}$
$=\frac{1}{8}\left\{3(1-\cos 2\theta )+2\sin \left(\frac{2\theta -4\theta }{2}\right).\sin \left(\frac{6\theta }{2}\right)\right\}$
$=\frac{1}{8}\{6{\sin }^2\theta -2\sin \theta .\sin 3\theta \}$
$=\frac{1}{8}\{6{\sin }^2\theta -2\sin \theta (3\sin \theta -4{\sin }^3\theta )\}$
$=\frac{1}{8}\{6{\sin }^2\theta -6{\sin }^2\theta +8{\sin }^4\theta \}={\sin }^4\theta$