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 $
• 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 $