Question

The value of $$\sin {\displaystyle \frac{\pi }{n}}+\sin {\displaystyle \frac{3\pi }{n}}+\sin {\displaystyle \frac{5\pi }{n}}+....$$ to n terms is equal to:
A). 1
B). 0
C). $${\displaystyle \frac{n}{2}}$$
D). None of these

Answer 1

Hint: The given series is of trigonometry identity in which “sin” is given with repeated angles, but having a numeric value multiplied by each angle, here to simplify the series we need to make the general equation for the series and after solving the general term here we can solve for the whole equation.
Formulae Used: Summation upto N terms of sin series, in which angles are in increasing order with fixed increment is given by:
$$\Rightarrow \sin (x)+\sin (x+y)+\sin (x+2y)+...+\sin (x(n-1)y)={\displaystyle \frac{\sin ({\displaystyle \frac{ny}{2}})}{\sin ({\displaystyle \frac{y}{2}})}}\sin \left(x+{\displaystyle \frac{n-1}{2}}y\right)$$

Complete step-by-step solution:
To find the general term for the given series here we need to make sure that how the numbers are given with the angles, on solving we get:
Every angle here is the odd multiple of the angle $${\displaystyle \frac{\pi }{n}}$$, and to write the general term for the odd number we use:
$$\Rightarrow oddmultiple=2n+1,n\in 0,1,2...$$
Summation of this sort of series for n terms is given by:
$$\Rightarrow \sin (x)+\sin (x+y)+\sin (x+2y)+...+\sin (x(n-1)y)={\displaystyle \frac{\sin ({\displaystyle \frac{ny}{2}})}{\sin ({\displaystyle \frac{y}{2}})}}\sin \left(x+{\displaystyle \frac{n-1}{2}}y\right)$$
Using this formulae in our series we have:
$$\begin{gathered} \Rightarrow \sin {\displaystyle \frac{\pi }{n}}+\sin {\displaystyle \frac{3\pi }{n}}+\sin {\displaystyle \frac{5\pi }{n}}+....+\sin \left({\displaystyle \frac{(2n+1)\pi }{n}}\right) \\ \Rightarrow {\displaystyle \frac{\sin \left({\displaystyle \frac{(2n)2\pi }{2n}}\right)}{\sin \left({\displaystyle \frac{2\pi }{2n}}\right)}}\sin \left({\displaystyle \frac{\pi }{n}}+n{\displaystyle \frac{2\pi }{n}}\right) \\ \Rightarrow {\displaystyle \frac{\sin \left(2\pi \right)}{\sin \left({\displaystyle \frac{2\pi }{2n}}\right)}}\sin \left({\displaystyle \frac{\pi }{n}}+n{\displaystyle \frac{2\pi }{n}}\right)=0(\sin ce\sin (2\pi )=0) \end{gathered}$$
Here we can see that the numerator has a term of “sin” function with angle of two pie, and “sin” gives the value of zero at this angle, hence the total term becomes zero here.

Note: Here in this kind of series question where trigonometric values and identities are used, we need to use the direct formulae here, as for this series we directly know the summation to n terms, hence put the value and get the correct answer.