Question

Consider the following statements
(1) lf $\sin A=\sin B$, then we have $\sin 2A=\sin 2B$ always
(2) The value of $\cos \frac{\pi }{7}\cos \frac{4\pi }{7}\cos \frac{5\pi }{7}$ is $\frac{1}{4}$.
Which of the statements given above is/are correct?

• A. Only (1)
• B. Only (2)
• C. Both (1) and (2)
• D. Neither (1) nor (2)

Answer 1

The correct option is D Neither (1) nor (2)

$\sin A=\sin B\cdots \left(1\right)$

$\sin 2A=\sin 2B$

$⟹2\sin A\cos A=2\sin B\cos B$

The above relation comes true only if $\cos A=\cos B$

But that may not happen at all cases.

$\cos \frac{\pi }{7}\cos \frac{4\pi }{7}\cos \frac{5\pi }{7}$

$⟹\frac{1}{2}\left(2\cos \frac{\pi }{7}\cos \frac{4\pi }{7}\right)\cos \frac{5\pi }{7}$

$⟹\frac{1}{2}(\cos \frac{5\pi }{7}+\cos \frac{3\pi }{7})\cos \frac{5\pi }{7}$

$⟹\frac{1}{2}{\cos }^2\frac{5\pi }{7}+\frac{1}{2}\left(\cos \frac{3\pi }{7}\cos \frac{5\pi }{7}\right)$

$⟹\frac{1}{4}(1+\cos \frac{10\pi }{7})+\frac{1}{4}(\cos \frac{8\pi }{7}+\cos \frac{2\pi }{7})$

$⟹\frac{1}{4}(1-\cos \frac{3\pi }{7}-\cos \frac{\pi }{7}+\cos \frac{2\pi }{7})$

$⟹\frac{1}{4}(1-2\cos \frac{2\pi }{7}\cos \frac{\pi }{7}+\cos \frac{2\pi }{7})$

This expression cannot be equal to $\frac{1}{4}$

So both the expressions are wrong.