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Prove That: Cos π 5 Cos 2 π 5 Cos 4 π 5 Cos 8 π 5 = − 1 16 - Mathematics

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Question

Prove that: \[\cos\frac{\pi}{5}\cos\frac{2\pi}{5}\cos\frac{4\pi}{5}\cos\frac{8\pi}{5} = \frac{- 1}{16}\]

 
Numerical
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Solution

\[\cos\frac{\pi}{5}\cos\frac{2\pi}{5}\cos\frac{4\pi}{5}\cos\frac{8\pi}{5}\]
\[ = \frac{1}{2\sin\frac{\pi}{5}}\left( 2\sin\frac{\pi}{5}\cos\frac{\pi}{5}\cos\frac{2\pi}{5}\cos\frac{4\pi}{5}\cos\frac{8\pi}{5} \right) \left( \text{ Multiplying and dividing by } \frac{1}{2\sin\frac{\pi}{5}} \right)\]
\[ = \frac{1}{2\sin\frac{\pi}{5}}\left( \sin\frac{2\pi}{5}\cos\frac{2\pi}{5}\cos\frac{4\pi}{5}\cos\frac{8\pi}{5} \right) \left( \sin2A = 2\text{ sin } A\text{ cos } A \right)\]
\[ = \frac{1}{4\sin\frac{\pi}{5}}\left( 2\sin\frac{2\pi}{5}\cos\frac{2\pi}{5}\cos\frac{4\pi}{5}\cos\frac{8\pi}{5} \right) \left( \text{ Multiplying and dividing by } 2 \right)\]

\[= \frac{1}{4\sin\frac{\pi}{5}}\left( \sin\frac{4\pi}{5}\cos\frac{4\pi}{5}\cos\frac{8\pi}{5} \right)\]
\[ = \frac{1}{8\sin\frac{\pi}{5}}\left( 2\sin\frac{4\pi}{5}\cos\frac{4\pi}{5}\cos\frac{8\pi}{5} \right) \left( \text{ Multiplying and dividing by 2 }  \right)\]
\[ = \frac{1}{8\sin\frac{\pi}{5}}\left( \sin\frac{8\pi}{5}\cos\frac{8\pi}{5} \right)\]
\[ = \frac{1}{16\sin\frac{\pi}{5}}\left( 2\sin\frac{8\pi}{5}\cos\frac{8\pi}{5} \right) \left(\text{  Multiplying and dividing by 2 }  \right)\]

\[= \frac{\sin\frac{16\pi}{5}}{16\sin\frac{\pi}{5}}\]

\[ = \frac{\sin\left( 3\pi + \frac{\pi}{5} \right)}{16\sin\frac{\pi}{5}}\]

\[ = \frac{- \sin\frac{\pi}{5}}{16\sin\frac{\pi}{5}} \left[ \sin\left( 3\pi + \theta \right) = - \sin\theta \right]\]

\[ = \frac{- 1}{16}\]

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Values of Trigonometric Functions at Multiples and Submultiples of an Angle
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Chapter 9: Values of Trigonometric function at multiples and submultiples of an angle - Exercise 9.1 [Page 29]

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RD Sharma Mathematics [English] Class 11
Chapter 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.1 | Q 35 | Page 29

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