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Prove That: Sin 2 π 18 + Sin 2 π 9 + Sin 2 7 π 18 + Sin 2 4 π 9 = 2 - Mathematics

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प्रश्न

Prove that:
\[\sin^2 \frac{\pi}{18} + \sin^2 \frac{\pi}{9} + \sin^2 \frac{7\pi}{18} + \sin^2 \frac{4\pi}{9} = 2\]

 
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उत्तर

LHS = \[\sin^2 \frac{\pi}{18} + \sin^2 \frac{\pi}{9} + \sin^2 \frac{7\pi}{18} + \sin^2 \frac{4\pi}{9}\]
\[ = \sin^2 \frac{\pi}{18} + \sin^2 \frac{2\pi}{18} + \sin^2 \frac{7\pi}{18} + \sin^2 \frac{8\pi}{18}\]
\[ = \sin^2 \frac{\pi}{18} + \sin^2 \frac{2\pi}{18} + \sin^2 \left( \frac{7\pi}{18} \right) + \sin^2 \left( \frac{8\pi}{18} \right)\]
\[ = \sin^2 \frac{\pi}{18} + \sin^2 \frac{2\pi}{18} + \sin^2 \left( \frac{\pi}{2} - \frac{2\pi}{18} \right) + \sin^2 \left( \frac{\pi}{2} - \frac{\pi}{18} \right)\]
\[ = \sin^2 \frac{\pi}{18} + \sin^2 \frac{2\pi}{18} + \cos^2 \frac{2\pi}{18} + \cos^2 \frac{\pi}{18}\]
\[ = \sin^2 \frac{\pi}{18} + \cos^2 \frac{\pi}{18} + \sin^2 \frac{2\pi}{18} + \cos^2 \frac{2\pi}{18}\]
\[ = 1 + 1\]
\[ = 2\]
 = RHS
Hence proved.

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पाठ 5: Trigonometric Functions - Exercise 5.3 [पृष्ठ ४०]

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आरडी शर्मा Mathematics [English] Class 11
पाठ 5 Trigonometric Functions
Exercise 5.3 | Q 4 | पृष्ठ ४०

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