Find the value of sin2π/10 + sin24π/10 + sin26π/10 + sin29π/10.

1.	Find the value of sin2π/10 + sin24π/10 + sin26π/10 + sin29π/10.
Answer» The answer is 0 sin²\(\frac{\pi}{10}\) + sin²\(\frac{4\pi}{10}\) + sin²\(\frac{6\pi}{10}\) + sin²\(\frac{9\pi}{10}\) = sin²\(\frac{\pi}{10}\) + sin²\(\frac{4\pi}{10}\) + sin²(\(\pi-\frac{4\pi}{10}\)) + sin²(\(\pi-\frac{\pi}{10}\)) = sin²\(\frac{\pi}{10}\) + sin²\(\frac{4\pi}{10}\) + sin²\(\frac{6\pi}{10}\) + sin²\(\frac{\pi}{10}\) = 2 ( sin²\(\frac{\pi}{10}\) + sin²\(\frac{4\pi}{10}\) ) = 2(sin²(\(\frac{\pi}{10}\)) + sin²(\(\frac\pi2-\frac\pi{10}\))) = 2 (sin²\(\frac{\pi}{10}\) + cos²\(\frac{\pi}{10}\)) = 2 x 1 = 2 (\(\because\) sin²θ + cos²θ = 1)

Answer»

The answer is 0

sin²\(\frac{\pi}{10}\) + sin²\(\frac{4\pi}{10}\) + sin²\(\frac{6\pi}{10}\) + sin²\(\frac{9\pi}{10}\)

= sin²\(\frac{\pi}{10}\) + sin²\(\frac{4\pi}{10}\) + sin²(\(\pi-\frac{4\pi}{10}\)) + sin²(\(\pi-\frac{\pi}{10}\))

= sin²\(\frac{\pi}{10}\) + sin²\(\frac{4\pi}{10}\) + sin²\(\frac{6\pi}{10}\) + sin²\(\frac{\pi}{10}\)

= 2 ( sin²\(\frac{\pi}{10}\) + sin²\(\frac{4\pi}{10}\) )

= 2(sin²(\(\frac{\pi}{10}\)) + sin²(\(\frac\pi2-\frac\pi{10}\)))

= 2 (sin²\(\frac{\pi}{10}\) + cos²\(\frac{\pi}{10}\))

= 2 x 1 = 2

(\(\because\) sin²θ + cos²θ = 1)

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