# DIV » DIVProve That: Sin ( π 3 − X ) Cos ( π 6 + X ) + Cos ( π 3 − X ) Sin ( π 6 + X ) = 1 - Mathematics

Prove that:

$\sin\left( \frac{\pi}{3} - x \right)\cos\left( \frac{\pi}{6} + x \right) + \cos\left( \frac{\pi}{3} - x \right)\sin\left( \frac{\pi}{6} + x \right) = 1$

#### Solution

$\frac{\pi}{3} = 60^\circ, \frac{\pi}{6} = 30^\circ$
$\text{ LHS }= \sin\left( 60^\circ - x \right) \cos\left( 30^\circ + x \right) + \cos\left( 60^\circ - x \right) \sin\left( 30^\circ + x \right)$
$= \sin\left[ \left( 60^\circ - x \right) + \left( 30^\circ + x \right) \right] (\text{ Using the formula }\sin A \cos B + \cos A \sin B = \sin\left( A + B \right)$
$\text{ and taking }A = 60^\circ - x\text{ and }B = 30^\circ + x$
$= \sin90^\circ$
$= 1$
= RHS
Hence proved.

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 7 Values of Trigonometric function at sum or difference of angles
Exercise 7.1 | Q 12.1 | Page 19