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Prove That: Sin 2 π 8 + Sin 2 3 π 8 + Sin 2 5 π 8 + Sin 2 7 π 8 = 2 - Mathematics

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Question

Prove that: \[\sin^2 \frac{\pi}{8} + \sin^2 \frac{3\pi}{8} + \sin^2 \frac{5\pi}{8} + \sin^2 \frac{7\pi}{8} = 2\]

Numerical
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Solution

\[LHS = \sin^2 \frac{\pi}{8} + \sin^2 \frac{3\pi}{8} + \sin^2 \frac{5\pi}{8} + \sin^2 \frac{7\pi}{8}\]

\[ = \sin^2 \left( \frac{\pi}{2} - \frac{3\pi}{8} \right) + \sin^2 \left( \frac{\pi}{2} - \frac{\pi}{8} \right) + \sin^2 \frac{5\pi}{8} + \sin^2 \frac{7\pi}{8}\]

\[ = \cos^2 \frac{3\pi}{8} + \sin^2 \frac{\pi}{8} + \sin^2 \left( \pi - \frac{3\pi}{8} \right) + \sin^2 \left( \pi - \frac{\pi}{8} \right)\]

\[= \cos^2 \frac{3\pi}{8} + \sin^2 \frac{\pi}{8} + \sin^2 \frac{3\pi}{8} + \cos^2 \frac{\pi}{8}\]

\[ = \left( \cos^2 \frac{\pi}{8} + \sin^2 \frac{\pi}{8} \right) + \left( \cos^2 \frac{3\pi}{8} + \sin^2 \frac{3\pi}{8} \right)\]

\[ = 1 + 1 = 2 = RHS\]

\[\text{ Hence proved }  .\]

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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 28]

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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 10 | Page 28

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