MCQ
The period of ${\sin ^4}x + {\cos ^4}x $ is
- ✓$\pi /2$
- B$\pi $
- C$2\pi $
- D$3\pi /2$
$= {({\sin ^2}x + {\cos ^2}x)^2} - 2{\sin ^2}x{\cos ^2}x$
$= 1 - \frac{{4{{\sin }^2}x{{\cos }^2}x}}{2} = 1 - \frac{{{{\sin }^2}2x}}{2}$
$=1 - \frac{1}{4}(2{\sin ^2}2x) = 1 - \left( {\frac{{1 - \cos x}}{4}} \right)$
$ = \frac{3}{4} + \frac{1}{4}\cos 4x$
Hence the period of function = $\frac{{2\pi }}{4} = \frac{\pi }{2}$.
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