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

Answer in Brief

Prove that:

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

 

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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
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