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Question

Mathematics Question on Inverse Trigonometric Functions

The period of sin⁡4 x+cos⁡4 x\sin^4 \, x + \cos^4 \, x is

A

π42\frac{\pi^4}{2}

B

π22\frac{\pi^2}{2}

C

π4\frac{\pi}{4}

D

π2\frac{\pi}{2}

Answer

π2\frac{\pi}{2}

Explanation

Solution

Let sin⁡4 x+cos⁡4 x\sin^4 \, x + \cos^4 \, x
=  (sin⁡2 x+cos⁡2 x)2−2 sin⁡2 x cos⁡2 x= \, \, (\sin^2 \, x + \cos^2 \, x )^2 - 2 \, \sin^2 \, x \, \cos^2 \, x
=  1−14.2 sin⁡ 2x2= \, \, 1- \frac{1}{4} . 2 \, \sin \, 2x^2
=  1−14  1−cos⁡ 4x= \, \, 1 - \frac{1}{4} \, \, 1- \cos \, 4x
=34+cos⁡ 4x4= \frac{3}{4} + \frac{\cos \, 4x}{4}
∴\therefore Period of f(x)=2π4=π2f(x) = \frac{2 \pi}{4} = \frac{\pi}{2}