Question

The sum $\cos \frac{\pi }{9}+\cos \frac{2\pi }{9}+\cos \frac{3\pi }{9}+.......+\cos \frac{17\pi }{9}$ equal ?

• A. $\frac{1}{2}$
• B. $-\frac{1}{2}$
• C. $1$
• D. $-1$

Answer 1

Answer: (D)

$-1$

Formula: $\cos \left(\theta \right)+\cos (\theta +\varphi )+\cos (\theta =2\varphi )+.....\cos (\theta +(n-1\left)\varphi \right)=\frac{\sin \left(\frac{n\varphi }{2}\right)}{\sin \left(\frac{\varphi }{2}\right)}\cdot \cos (\theta +(n-1\left)\frac{\varphi }{2}\right)$

Here, $\theta =p=\frac{\pi }{9};n=17$

$\therefore$ sum $=\frac{\sin \left(17\frac{\pi }{18}\right)}{\sin \left(\frac{\pi }{18}\right)}\cdot \cos (\frac{\pi }{9}+(16\left)\frac{\pi }{18}\right)$

$=\frac{\sin (\pi \cdot \frac{\pi }{18})}{\sin \left(\frac{\pi }{18}\right)}\cdot \cos \left(\pi \right)$ $\{\because \sin (\pi -\theta )=\sin (\theta \left)\right\}$

$=\frac{\sin \left(\frac{\pi }{18}\right)}{\sin \left(\frac{\pi }{18}\right)}\cdot (-1)$

$=-1$.