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

The Correct Option is C

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 $

Answer 2

Given :
\(\frac{1}{8}(3-4\text{ }cos\text{ }2\theta +cos\text{ }4\theta )\)
As per formula : \(\cos\ 2A=2\cos^2A-1\)
Now, 
\(=\frac{1}{8}(3-4\cos2\theta+2\cos^22\theta-1)\)
\(=\frac{1}{8}(2\cos^22\theta-2.2\cos2\theta+2)\)
\(=\frac{2}{8}(\cos^22\theta-2\cos2\theta+1)\)
\(=\frac{1}{4}(\cos2\theta-1)^2 \ \ \ \ \ \ \ [\because a^2-2ab+b^2=(a-b)^2]\)
\(=\frac{1}{4}(1-2\sin^2\theta-1)^2\ \ \ \ \ \ \ [\because\cos2\theta=1-2\sin^2\theta]\)
\(=\frac{1}{4}(4\sin^2\theta)\)
⇒ \(\sin^4\theta\)
So, the correct option is (C) : $ {{\sin }^{4}}\theta $