Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# Prove That: Sin 10 π 3 Cos 13 π 6 + Cos 8 π 3 Sin 5 π 6 = − 1 - Mathematics

Prove that:

$\sin\frac{10\pi}{3}\cos\frac{13\pi}{6} + \cos\frac{8\pi}{3}\sin\frac{5\pi}{6} = - 1$

#### Solution

$\frac{10\pi}{3} = 600^\circ, \frac{13\pi}{6} = 390^\circ, \frac{8\pi}{3} = 480^\circ, \frac{5\pi}{6} = 150^\circ$

LHS = $\sin 600^\circ\cos 390^\circ + \cos 480^\circ \sin 150^\circ$

$= \sin \left( 90^\circ \times 6 + 60^\circ \right) \cos\left( 90^\circ \times 4 + 30^\circ \right) + \cos\left( 90^\circ \times 5 + 30^\circ \right) \sin\left( 90^\circ \times 1 + 60^\circ \right)$

$= \left[ - \sin 60^\circ \right] \cos30^\circ + \left[ - \sin 30^\circ \right] \cos 60^\circ$

$= - \sin 60^\circ \cos\left( 30^\circ \right) - \sin 30^\circ \cos 60^\circ$

$= - \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} - \frac{1}{2} \times \frac{1}{2}$

$= - \frac{3}{4} - \frac{1}{4}$

$= - 1$

= RHS

Hence proved.

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 5 Trigonometric Functions
Exercise 5.3 | Q 9.4 | Page 40