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Sin2 π/18 + Sin2 π/9 + Sin2 7π/18 + Sin2 4π/9 =

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प्रश्न

sin2 π/18 + sin2 π/9 + sin2 7π/18 + sin2 4π/9 =

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उत्तर

2
We have: 
\[ \sin^2 \frac{\pi}{18} + \sin^2 \frac{\pi}{9} + \sin^2 \frac{7\pi}{18} + \sin^2 \frac{4\pi}{9}\]
\[ = \sin^2 \frac{\pi}{18} + \sin^2 \frac{2\pi}{18} + \sin^2 \frac{7\pi}{18} + \sin^2 \frac{8\pi}{18}\]
\[ = \sin^2 \frac{\pi}{18} + \sin^2 \frac{2\pi}{18} + \sin^2 \left( \frac{7\pi}{18} \right) + \sin^2 \left( \frac{8\pi}{18} \right)\]
\[ = \sin^2 \frac{\pi}{18} + \sin^2 \frac{2\pi}{18} + \sin^2 \left( \frac{\pi}{2} - \frac{2\pi}{18} \right) + \sin^2 \left( \frac{\pi}{2} - \frac{\pi}{18} \right)\]
\[ = \sin^2 \frac{\pi}{18} + \sin^2 \frac{2\pi}{18} + \cos^2 \frac{2\pi}{18} + \cos^2 \frac{\pi}{18}\]
\[ = \sin^2 \frac{\pi}{18} + \cos^2 \frac{\pi}{18} + \sin^2 \frac{2\pi}{18} + \cos^2 \frac{2\pi}{18}\]
\[ = 1 + 1\]
\[ = 2\]

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अध्याय 5: Trigonometric Functions - Exercise 5.5 [पृष्ठ ४२]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 5 Trigonometric Functions
Exercise 5.5 | Q 17 | पृष्ठ ४२

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