Question

$f\left(x\right)=\sqrt{{(\cos x+\cos 2x+\cos 3x)}^2+{(\sin x+\sin 2x+\sin 3x)}^2}$ , then which of the following is / are correct

• A. $f\left(\frac{\pi }{12}\right)=\frac{\sqrt{1}+\sqrt{2}+\sqrt{3}}{\sqrt{2}}$
• B. $f\left(\frac{\pi }{5}\right)=\frac{3+\sqrt{5}}{2}$
• C. $f\left(x\right)isanevenfunction$
• D. None of these

Answer 1

Answer: (C)

$f\left(x\right)isanevenfunction$

Given

$f\left(x\right)=\sqrt{{(\cos x+\cos 2x+\cos 3x)}^2+{(\sin x+\sin 2x+\sin 3x)}^2}$

Replace $x$ with $(-x)$

$⟹f(-x)=\sqrt{{\left(\cos \right(-x)+\cos (-2x)+\cos (-3x\left)\right)}^2+{\left(\sin \right(-x)+\sin (-2x)+\sin (-3x\left)\right)}^2}$

$⟹\sqrt{{(\cos x+\cos 2x+\cos 3x)}^2+{(-(\sin x+\sin 2x+\sin 3x\left)\right)}^2}$

$⟹\sqrt{{(\cos x+\cos 2x+\cos 3x)}^2+{(\sin x+\sin 2x+\sin 3x)}^2}$

$⟹f(-x)=f\left(x\right)$

$\therefore f\left(x\right)$ is an even function .