Question

$\cos \theta +\cos 2\theta +\cos 3\theta +\dots .+\cos \left\{\right(n-1\left)\theta \right\}+\cos n\theta$ is equal to

• A. $\sin \left(\frac{\theta }{2}\right)$
• B. $\frac{\cos (n+1)\theta }{\sin \left(\frac{\theta }{2}\right)}$
• C. $\frac{\cos (\theta +\frac{(n-1)\theta }{2})\sin \left(\frac{n\theta }{2}\right)}{\sin \frac{\theta }{2}}$
• D. $\sin \left(\frac{n\theta }{2}\right).\cos \left\{\right(n+1\left)\theta \right\}$

Answer 1

The correct option is C $\frac{\cos (\theta +\frac{(n-1)\theta }{2})\sin \left(\frac{n\theta }{2}\right)}{\sin \frac{\theta }{2}}$

$\Rightarrow$Using $\cos x+\cos (x+y)+\cos (x+2y)+...+\cos (x+(n-1)y)=\frac{\sin \frac{ny}{2}}{\sin \frac{y}{2}}\cos (x+\left(\frac{n-1}{2}\right)y)$

We get

$\Rightarrow$$\cos \theta +\cos 2\theta +\cos 3\theta +...+\cos n\theta$

$=\cos \left(\theta \right)+\cos (\theta +\theta )+\cos (\theta +2\theta )+...+\cos (\theta +(n-1)\theta )$

$=\frac{\sin \frac{n\theta }{2}}{\sin \frac{\theta }{2}}\cos (\theta +\left(\frac{n-1}{2}\right)\theta )$