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Question: \(\sin^{2}\frac{\pi}{8} + \sin^{2}\frac{3\pi}{8} + \sin^{2}\frac{5\pi}{8} + \sin^{2}\frac{7\pi}{8} =...

sin⁡2π8+sin⁡23π8+sin⁡25π8+sin⁡27π8=\sin^{2}\frac{\pi}{8} + \sin^{2}\frac{3\pi}{8} + \sin^{2}\frac{5\pi}{8} + \sin^{2}\frac{7\pi}{8} =

A

1

B

– 1

C

0

D

2

Answer

2

Explanation

Solution

sin⁡2π8+sin⁡23π8+sin⁡25π8+sin⁡27π8\sin^{2}\frac{\pi}{8} + \sin^{2}\frac{3\pi}{8} + \sin^{2}\frac{5\pi}{8} + \sin^{2}\frac{7\pi}{8}

=sin⁡2π8+sin⁡23π8+sin⁡23π8+sin⁡2π8= \sin^{2}\frac{\pi}{8} + \sin^{2}\frac{3\pi}{8} + \sin^{2}\frac{3\pi}{8} + \sin^{2}\frac{\pi}{8}

=2(sin⁡2π8+sin⁡23π8)=2×1=2= 2\left( \sin^{2}\frac{\pi}{8} + \sin^{2}\frac{3\pi}{8} \right) = 2 \times 1 = 2.