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Prove that: sin⁡π5sin⁡2π5sin⁡3π5sin⁡4π5=516

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प्रश्न

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

 
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उत्तर

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\[ = RHS\]

Thus, LHS = RHS
Hence,

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

 

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Values of Trigonometric Functions at Multiples and Submultiples of an Angle
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Values of Trigonometric function at multiples and submultiples of an angle - Exercise 9.3 [पृष्ठ ४२]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.3 | Q 9 | पृष्ठ ४२

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