ત્રિકોણમિતિય વિધયો — ગણિત ધોરણ 11 સાયન્સ — Question

$\sum_{r=1}^{n-1} \cos^2\frac {r\pi}{4}$

$= \frac{1}{2}\sum_{r=1}^{n-1} (1+\cos\frac{2r\pi}{4})$

$= \frac{1}{2} (n-1)+ \frac{1}{2}\left\{\cos\frac{2\pi} {n}+\cos\frac{4\pi}{n}+............+\cos \frac{(2n-2)\pi}{n}\right\} $

$=\frac{1}{2} (n-1) + \frac{1} {2} \frac{\sin(n-1) \frac{\pi}{n}}{\sin \frac{\pi}{n}} \cdot cos (\frac{2\pi}{n}+(n-2)\frac{\pi}{n})$

$=\frac{1}{2} (n-1)-\frac{1}{2}$

$\sum_{r=1}^{n-1} \cos^2\frac {r\pi}{4} =\frac{n}{2}-1 $