Question

The period of $si{n}^{4}x+co{s}^{4}x$ is.

• A. $\pi /2$
• B. $\pi$
• C. $2\pi$
• D. none of these

Answer 1

The correct option is A $\pi /2$

We have,
$\left(x\right)={\sin }^{4}x+{\cos }^{4}x$

$\Rightarrow \left(x\right)={\left({\sin }^{2}x\right)+\left({\cos }^{2}x\right)}^{2}2{\sin }^{2}xco{s}^{2}x$

$\Rightarrow \left(x\right)=({\sin }^{2}x+{\cos }^{2}x{)}^2-\frac{1}{2}(2\sin x\cos x{)}^2$

$\Rightarrow \left(x\right)=1-\frac{1}{2}\left\{si{n}^{2}2x\right\}$

$\Rightarrow \left(x\right)=1-\frac{1}{2}\left\{\frac{1-\cos 4x}{2}\right\}$

$\Rightarrow \left(x\right)=\frac{3}{4}-\frac{1}{4}\cos 4x$.

Since $\cos 4x$ is a periodic function with period $\frac{2\pi }{4}=\frac{\pi }{2}$

So, (x) is periodic with period $\frac{\pi }{2}$.