Prove That: Sin 2 ( π 8 + X 2 ) − Sin 2 ( π 8 − X 2 ) = 1 √ 2 Sin X - Mathematics

Numerical

Prove that:  $\sin^2 \left( \frac{\pi}{8} + \frac{x}{2} \right) - \sin^2 \left( \frac{\pi}{8} - \frac{x}{2} \right) = \frac{1}{\sqrt{2}} \sin x$

Solution

$LHS = \sin^2 \left( \frac{\pi}{8} + \frac{x}{2} \right) - \sin^2 \left( \frac{\pi}{8} - \frac{x}{2} \right)$

$= \frac{1}{2}\left\{ 1 - \cos2\left( \frac{\pi}{8} + \frac{x}{2} \right) \right\} - \frac{1}{2}\left\{ 1 - \cos2\left( \frac{\pi}{8} - \frac{x}{2} \right) \right\}$

$= \frac{1}{2}\left\{ \cos\left( \frac{\pi}{4} - x \right) - \cos\left( \frac{\pi}{4} + x \right) \right\}$

Using the identit

$\text{ cos } C - \text{ cos } D = - 2\sin\frac{C + D}{2}\sin\frac{C - D}{2}$ , we get

$= \frac{1}{2}\left\{ - 2\sin\left( \frac{\left( \frac{\pi}{4} - x \right) + \left( \frac{\pi}{4} + x \right)}{2} \right)\sin\left( \frac{\left( \frac{\pi}{4} - x \right) - \left( \frac{\pi}{4} + x \right)}{2} \right) \right\}$

$= - \sin\frac{\pi}{4}\sin\left( - x \right)$

$= \sin\frac{\pi}{4}\text{ sin } x \left[ \because \sin\left( - x \right) = - \text{ sin } x \right]$

$= \frac{1}{\sqrt{2}}\text{ sin } x = RHS$

$\text{ Hence proved } .$

Concept: Values of Trigonometric Functions at Multiples and Submultiples of an Angle
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.1 | Q 12 | Page 28