Question

Find the value of $si{n}^2\frac{\pi }{16}$ + $si{n}^2\frac{2\pi }{16}$ + $si{n}^2\frac{3\pi }{16}$ +.....$si{n}^2\frac{16\pi }{16}$

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Answer 1

Angles in the sine function are in arithmetic progression we can use $\frac{sine}{cosine}$ series.

We know cos 2$\theta$ = 1 - 2 $si{n}^2\theta$

$si{n}^2\theta$ = $\frac{1}{2}$ (1 - $cos2\theta$)

Each term of the given series can be written as $cos2\theta$ from

$\frac{1}{2}(1-cos\frac{2\pi }{16})$ + $\frac{1}{2}(1-cos\frac{4\pi }{16})$ + ....$\frac{1}{2}(1-cos\frac{32\pi }{16})$

= $\frac{16}{2}$ - $\frac{1}{2}[cos\frac{2\pi }{16}+cos\frac{4\pi }{16}+cos\frac{6\pi }{16}+....cos\frac{32\pi }{16}]$

$\alpha =\frac{2\pi }{16}$,$\beta =\frac{2\pi }{16}$, n = 16

= $\frac{16}{2}$ - $\frac{1}{2}[\frac{sin16\times \frac{2\pi }{16\times 2}}{sin\left(\frac{2\pi }{16\times 2}\right)}.cos(\frac{2\pi }{16}+\frac{15.2\pi }{16\times 2})]$

= 8 - $\frac{1}{2}[\frac{sin\pi }{sin\left(\frac{\pi }{16}\right)}.cos\left(\frac{17\pi }{16}\right)]$

= 8 - $\frac{1}{2}\times 0$

= 8