Question

${\cos }^4\theta =\frac{1}{8}\cos 4\theta +\frac{1}{2}\cos 2\theta +\frac{3}{8}$

Answer 1

R.H.S$=\frac{1}{8}\cos 4\theta +\frac{1}{2}\cos 2\theta +\frac{3}{8}$

$=\frac{1}{8}(2{\cos }^{2}2\theta -1)+\frac{1}{2}(2{\cos }^2\theta -1)+\frac{3}{8}$

$=\frac{1}{4}{\cos }^{2}2\theta -\frac{1}{8}+{\cos }^2\theta -\frac{1}{2}+\frac{3}{8}$

$=\frac{1}{4}{(2{\cos }^2\theta -1)}^2+{\cos }^2\theta -\frac{1}{2}+\frac{3-1}{8}$

$=\frac{1}{4}(4{\cos }^4\theta +1-4{\cos }^2\theta )+{\cos }^2\theta -\frac{1}{2}+\frac{2}{8}$

$={\cos }^4\theta +\frac{1}{4}-{\cos }^2\theta +{\cos }^2\theta -\frac{1}{4}$

$={\cos }^4\theta$

$=$L.H.S

Hence proved