Question

Let $y=\sqrt{(\sin x+\sin 2x+\sin 3x{)}^2+(\cos x+\cos 2x+\cos 3x{)}^2}$ then which of the following(s) is correct?

• A. $\frac{dy}{dx}$ when $x=\frac{\pi }{2}$ is $-2$
• B. Value of y when $x=\frac{\pi }{5}$ is $\frac{3+\sqrt{5}}{2}$
• C. Value of y when $x=\frac{\pi }{12}$ is $\frac{\sqrt{1}+\sqrt{2}+\sqrt{3}}{2}$
• D. y simplifies to $(1+2\cos x)$ in $[0,\pi ]$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $\frac{dy}{dx}$ when $x=\frac{\pi }{2}$ is $-2$
B Value of y when $x=\frac{\pi }{5}$ is $\frac{3+\sqrt{5}}{2}$
C Value of y when $x=\frac{\pi }{12}$ is $\frac{\sqrt{1}+\sqrt{2}+\sqrt{3}}{2}$

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

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

$=\sqrt{1+1+1+2\cos (2x-x)+2\cos (3x-2x)+2\cos (3x-x)}=\sqrt{3+4\cos x+2\cos 2x}=\sqrt{4{\cos }^{2}x+4\cos x+1}$

$=|2\cos x+1|$

At $x=\frac{\pi }{2},\frac{dy}{dx}=-2\sin x=-2\sin \pi /2=-2$

At $x=\frac{\pi }{5},y=2\cos \frac{\pi }{5}+1=2\times \frac{\sqrt{5}+1}{4}+1=\frac{\sqrt{5}+3}{2}$

At $x=\frac{\pi }{12},y=2\cos \frac{\pi }{12}+1=2\times \frac{\sqrt{3}+1}{2\sqrt{2}}+1=\frac{\sqrt{1}+\sqrt{2}+\sqrt{3}}{\sqrt{2}}$