NIMCET 2020 — Mathematics PYQ

Concept:

Trigonometric Identities:

- $\sin 2\theta =2\sin \theta \cos \theta$

Trigonometric Ratios for Allied Angles:

- $\sin (-\theta )=-\sin \theta$
- $\cos (-\theta )=\cos \theta$
- $\sin (n\pi +\theta )=(-1{)}^n\sin \theta$
- $\cos (n\pi +\theta )=(-1{)}^n\cos \theta$
- $\sin \left[(2n+1)\frac{\pi }{2}+\theta \right]=(-1{)}^n\cos \theta$
- $\cos \left[(2n+1)\frac{\pi }{2}+\theta \right]=(-1{)}^n(-\sin \theta )$

Calculation :

$\sin 10^{\circ }\sin 50^{\circ }\sin 70^{\circ }$

$=\sin 10^{\circ }\sin (90^{\circ }-40^{\circ })\sin (90^{\circ }-20^{\circ })$

$=\sin 10^{\circ }\cos 40^{\circ }\cos 20^{\circ }$

$=\left(\frac{2\sin 10^{\circ }\cos 10^{\circ }}{2\cos 10^{\circ }}\right)\cos 40^{\circ }\cos 20^{\circ }$

$=\left(\frac{\sin 20^{\circ }}{2\cos 10^{\circ }}\right)\cos 40^{\circ }\cos 20^{\circ }$

$=\left(\frac{2\sin 20^{\circ }\cos 20^{\circ }}{4\cos 10^{\circ }}\right)\cos 40^{\circ }$

$=\frac{2\sin 40^{\circ }\cos 40^{\circ }}{8\cos 10^{\circ }}$

$=\frac{\sin 80^{\circ }}{8\cos 10^{\circ }}$

$=\frac{\sin (90^{\circ }-10^{\circ })}{8\cos 10^{\circ }}$

$=\frac{\cos 10^{\circ }}{8\cos 10^{\circ }}$

$=\frac{1}{8}$