Question

If $B$ the exterior angle of a regular polygon of $n-sides$ and $A$ is any constant then $\cos A+\cos (A+B)+\cos (A+2B)+....n$ terms is equal to:

• A. $0$
• B. $\cos A$
• C. $1$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

The correct option is A $0$

The sum of exterior angles of a polygon is $2\pi$ or ${360}^{\circ }$

If it is a regular polygon

Exterior Angles $=\frac{2\pi }{n}$

n is no of sides

According to question

$B=\frac{2\pi }{n}.$

$\cos A+\cos (A+B)+\cos (A+2B)+...n+terms$

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

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

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

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

$\{\because \cos (2\pi -\theta )=\cos \theta \}$

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

$\sin \frac{4\pi }{n}+...+\cos A.\cos \frac{4\pi }{n}+\sin A\sin \frac{4\pi }{n}+\cos A.\cos \frac{2\pi }{n}+\sin A.\sin \frac{\pi }{n}$

$=\cos A+\cos A.\cos \frac{2\pi }{n}+\cos A\cos \frac{4\pi }{n}+...\cos A\cos \frac{4\pi }{n}+\cos A.\cos \frac{2\pi }{n}.$

$=\cos A\{1+\cos \frac{2\pi }{n}+\cos \frac{4\pi }{n}+...\cos \frac{4\pi }{n}+\cos \frac{2\pi }{n}\}$

$=\cos A(1-1)$

$=0$