Question

Let $y=\frac{\cos x+\cos 2x+\cos 3x+\cos 4x+\cos 5x+\cos 6x+\cos 7x}{\sin x+\sin 2x+\sin 3x+\sin 4x+\sin 5x+\sin 6x+\sin 7x},$ then which of the following hold good?

• A. The value of y when $x=\pi /8$ is not defined
• B. The value of y when $x=\pi /16$ is 1.
• C. The value of y when $x=\pi /32$ is $\sqrt{2}-1$ .
• D. The value of y when $x=\pi /48$ is $2+\sqrt{3}$ .

Answer 1

Answer: (B), (D)

The correct options are
B The value of y when $x=\pi /16$ is 1.
D The value of y when $x=\pi /48$ is $2+\sqrt{3}$ .

$\sin x+\sin 2x+\sin 3x+\sin 4x+\sin 5x+\sin 6x+\sin 7x=\frac{\sin \frac{7x}{2}}{\sin \frac{x}{2}}\sin 4x$

$\cos x+\cos 2x+\cos 3x+\cos 4x+\cos 5x+\cos 6x+\cos 7x=\frac{\sin \frac{7x}{2}}{\sin \frac{x}{2}}\cos 4x$

$⟹y=\frac{\cos x+\cos 2x+\cos 3x+\cos 4x+\cos 5x+\cos 6x+\cos 7x}{\sin x+\sin 2x+\sin 3x+\sin 4x+\sin 5x+\sin 6x+\sin 7x}=\cot 4x$

$x=\frac{\pi }{8}⟹y=0$

$x=\frac{\pi }{16}⟹y=\cot \frac{\pi }{4}=1$

$x=\frac{\pi }{32}⟹y=\cot \frac{\pi }{8}=\sqrt{2}+1$

$x=\frac{\pi }{48}⟹y=\cot \frac{\pi }{12}=2+\sqrt{3}$