# Prove that : Sin π 5 Sin 2 π 5 Sin 3 π 5 Sin 4 π 5 = 5 16 - Mathematics

Numerical

Prove that : $\sin\frac{\pi}{5}\sin\frac{2\pi}{5}\sin\frac{3\pi}{5}\sin\frac{4\pi}{5} = \frac{5}{16}$

#### Solution

$LHS = \sin\frac{\pi}{5}\sin\frac{2\pi}{5}\sin\frac{3\pi}{5}\sin\frac{4\pi}{5}$

$= \frac{1}{2}\left( 2 \sin\frac{\pi}{5} \sin\frac{4\pi}{5} \right)\frac{1}{2}\left( 2 \sin\frac{2\pi}{5} \sin\frac{3\pi}{5} \right)$

$= \frac{1}{4}\left( \cos\left( \frac{\pi}{5} - \frac{4\pi}{5} \right) - \cos\left( \frac{\pi}{5} + \frac{4\pi}{5} \right) \right)\left( \cos\left( \frac{2\pi}{5} - \frac{3\pi}{5} \right) - \cos\left( \frac{2\pi}{5} + \frac{3\pi}{5} \right) \right)$

$= \frac{1}{4}\left( \cos\left( \frac{- 3\pi}{5} \right) - \cos\left( \frac{5\pi}{5} \right) \right)\left( \cos\left( \frac{- \pi}{5} \right) - \cos\left( \frac{5\pi}{5} \right) \right)$

$= \frac{1}{4}\left( \cos\left( \frac{3\pi}{5} \right) - \cos\left( \pi \right) \right)\left( \cos\left( \frac{\pi}{5} \right) - \cos\left( \pi \right) \right)$

$= \frac{1}{4}\left( \cos\left( \frac{3\pi}{5} \right) + 1 \right)\left( \cos\left( \frac{\pi}{5} \right) + 1 \right)$

$= \frac{1}{4}\left( \cos\left( \pi - \frac{2\pi}{5} \right) + 1 \right)\left( \cos\left( \frac{\pi}{5} \right) + 1 \right)$

$= \frac{1}{4}\left( - \cos\left( \frac{2\pi}{5} \right) + 1 \right)\left( \left( \frac{\sqrt{5} + 1}{4} \right) + 1 \right) \left( \because \cos \frac{\pi}{5} = \frac{\sqrt{5} + 1}{4} \right)$

$= \frac{1}{4}\left( - \left( \frac{\sqrt{5} - 1}{4} \right) + 1 \right)\left( \left( \frac{\sqrt{5} + 1}{4} \right) + 1 \right) \left( \because \cos \frac{2\pi}{5} = \frac{\sqrt{5} - 1}{4} \right)$

$= \frac{1}{4}\left( - \left( \frac{\sqrt{5} - 1}{4} \right)\left( \frac{\sqrt{5} + 1}{4} \right) - \left( \frac{\sqrt{5} - 1}{4} \right) + \left( \frac{\sqrt{5} + 1}{4} \right) + 1 \right)$

$= \frac{1}{4}\left( - \left( \frac{\left( \sqrt{5} \right)^2 - 1}{16} \right) + \left( \frac{\sqrt{5} + 1 - \sqrt{5} + 1}{4} \right) + 1 \right)$

$= \frac{1}{4}\left( - \left( \frac{4}{16} \right) + \left( \frac{2}{4} \right) + 1 \right)$

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

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

$= \frac{5}{16}$

$= RHS$

Thus, LHS = RHS
Hence,

$\sin\frac{\pi}{5}\sin\frac{2\pi}{5}\sin\frac{3\pi}{5}\sin\frac{4\pi}{5} = \frac{5}{16}$

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 9.3 | Q 9 | Page 42