Question

1. $\cos \frac{2\pi }{15}\cos \frac{4\pi }{15}\cos \frac{8\pi }{15}\cos \frac{16\pi }{15}=\frac{1}{16}$.
2. One value of $A$ which satisfies the equation ${\sin }^{4}A-2{\sin }^{2}A-1$ lies between 0 and $2\pi$.

If 1,2 are true then enter 1 else 0.

Answer 1

State which of the following true.

$\frac{\cos 2\pi }{15}\frac{\cos 4\pi }{15}\frac{\cos 8\pi }{15}\cos (\pi +\frac{\pi }{15})$

$\frac{\cos \pi }{2}\frac{\cos 2\pi }{15}\frac{\cos 4\pi }{15}\frac{\cos 8\pi }{15}$ $\left[\cos \right(\pi +\theta )=-\cos \theta ]$

$\left[\begin{array}{c}\because \cos A\cos 2A\cos 2^{2}A\cos 2^{3}A....\cos 2^{n-1}A\\ =\frac{\sin 2^{n}A}{2n\sin A}\end{array}\right]$

if $A=\frac{\pi }{15}$

then

$=-\cos A\cos 2A\cos 2^{2}A\cos 2^3$ $\left[\begin{array}{c}\because n-1=3\\ \therefore n=4\end{array}\right]$

$=-\frac{-\sin 2^{4}A}{2^4\sin A}-\frac{-\sin 16\times \frac{\pi }{15}}{16.\sin \frac{\pi }{15}}$

$=-\left(\frac{-sin\frac{\pi }{15}}{16\sin \frac{\pi }{15}}\right)$ $(\frac{16\pi }{15}=(\pi +\frac{\pi }{15}\left)\right)$

This statement is true $=\frac{1}{16}H.P$ $[\because \sin (\pi +\theta )=-\sin \theta ]$

(ii)

${\sin }^{4}A-2{\sin }^{2}A-1=0$ lies between $0$ to $2\pi$

$\Rightarrow {\sin }^{4}A-2{\sin }^{2}A+1-2=0$

$\Rightarrow ({\sin }^{2}A-1{)}^2-2=0$.........(1)

$\because 0\le {\sin }^{2}A\le 1$

$\Rightarrow -1\le {\sin }^{2}A-1\le 0$

$\Rightarrow 0\le ({\sin }^{2}A-1{)}^2\le 1$

$\Rightarrow -2\le ({\sin }^{2}A-1{)}^2-2\le -1$.........(1)

from eq (1) and (2)

$({\sin }^{2}A-1{)}^2-2$ does not have a solution.

So, statement (ii) is false.

So, Ans (1) is right