Question

$\left(\sin \frac{\pi }{10}+\sin \frac{13\pi }{10}\right)\left({\cos }^2\frac{\pi }{6}-{\cos }^2\frac{\pi }{10}\right)$ is equal to.

• A. $\frac{\sqrt{5}-1}{8}$
• B. $-\frac{1}{2}$
• C. $\frac{\sqrt{5}-1}{16}$
• D. None of these

Answer 1

The correct option is C $\frac{\sqrt{5}-1}{16}$

Let $A=\frac{\pi }{10}$

$\therefore 5A=\frac{\pi }{2}$

$\therefore 3A=\frac{\pi }{2}-2A$

$\therefore \sin 3A=\sin (\frac{\pi }{2}-2A)$

$\therefore 3\sin A-4{\sin }^{3}A=\cos 2A$

$\therefore 4{\sin }^{3}A-3\sin A-2{\sin }^{2}A+1=0$

$\therefore (\sin A-1)(4{\sin }^{2}A+2\sin A-1)=0$

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

$\therefore \sin A=\frac{-1\pm \sqrt{5}}{4}$

but $\sin A$ cannot be negative as it lies in $[0,\frac{\pi }{2}]$.

$\therefore \sin A=\frac{\sqrt{5}-1}{4}$

$\therefore \cos A=\frac{\sqrt{10+2\sqrt{5}}}{4}$

Now,

$\sin \frac{13\pi }{10}=\sin (\pi +3A)$

$\therefore \sin \frac{13\pi }{10}=-\sin 3A$

$\therefore \sin \frac{13\pi }{10}=4{\sin }^{3}A-3\sin A$

$\therefore \sin \frac{13\pi }{10}=\sin A(\frac{(\sqrt{5}-1{)}^2}{4}-3)$

$\therefore \sin \frac{13\pi }{10}=\sin A\frac{5+1-2\sqrt{5}-12}{4}$

$\therefore \sin \frac{13\pi }{10}=\frac{\sqrt{5}-1}{4}\frac{-(\sqrt{5}+1{)}^2}{4}$

$\therefore \sin \frac{13\pi }{10}=-\frac{4(\sqrt{5}+1)}{16}$

$\therefore \sin \frac{13\pi }{10}=-\frac{\sqrt{5}+1}{4}$

Hence,

$(\sin A+\sin 13A)({\cos }^2\frac{\pi }{6}-{\cos }^{2}A)=(\frac{\sqrt{5}-1}{4}-\frac{\sqrt{5}+1}{4})(\frac{3}{4}-\frac{(10+2\sqrt{5})}{16})$

$=(-\frac{2}{4})\left(\frac{(2-2\sqrt{5})}{16}\right)$

$=\left(\frac{\sqrt{5}-1}{16}\right)$

This is the required answer.