Question

The period of $\sin^{4}x + \cos^{4}x$ is

• A. $\frac{\pi}{2}$
• B. $\pi$
• C. $2\pi$
• D. $\frac{3\pi}{2}$

Answer 1

Answer: (A)

$\frac{\pi}{2}$

Answer 2

$(\sin^{2}x)^{2} + (\cos^{2}x)^{2} = 1 - 2\sin^{2} ⥂ x\cos^{2}x = 1 - \frac{1}{4}2\sin^{2}2x = 1 - \left( \frac{1 - \cos 4x}{4} \right) = \frac{3}{4} + \frac{1}{4}\cos 4x$Therefore period is $\frac{2\pi}{4} = \frac{\pi}{2}.$

Trick: $\mathbf{f(x) =}\mathbf{\sin}^{\mathbf{4}}\mathbf{x}\mathbf{+}\mathbf{\cos}^{\mathbf{4}}\mathbf{x}$

⇒$f\left( \frac{\pi}{2} + x \right) = \sin^{4}\left( \frac{\pi}{2} + x \right) + \cos^{4}\left( \frac{\pi}{2} + x \right)$

⇒$\mathbf{f}\left( \frac{\mathbf{\pi}}{\mathbf{2}}\mathbf{+ x} \right)\mathbf{=}\mathbf{\cos}^{\mathbf{4}}\mathbf{x}\mathbf{+}\mathbf{\sin}^{\mathbf{4}}\mathbf{x}\mathbf{= f(x)}$ Hence the period is $\frac{\pi}{2}.$