Question

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

Answer 1

Hint: For answering this question we will use the formulae $\sin \left( x \right)+\sin \left( x+y \right)+\sin \left( x+2y \right)+.............+\sin \left( x+\left( N-1 \right)y \right)=\dfrac{\sin \left( \dfrac{Ny}{2} \right)}{\sin \left( \dfrac{y}{2} \right)}\sin \left( x+\dfrac{N-1}{2}y \right)$
which we have learnt from the basic concept and simplify the given expression $\sin \dfrac{\pi }{n}+\sin \dfrac{3\pi }{n}+\sin \dfrac{5\pi }{n}+.............\text{to n terms}$ and find its value.

Complete step-by-step solution
Now considering from the question we have the expression $\sin \dfrac{\pi }{n}+\sin \dfrac{3\pi }{n}+\sin \dfrac{5\pi }{n}+.............\text{to n terms}$ .
We will use the formulae learnt from the basic concept stated as $\sin \left( x \right)+\sin \left( x+y \right)+\sin \left( x+2y \right)+.............+\sin \left( x+\left( N-1 \right)y \right)=\dfrac{\sin \left( \dfrac{Ny}{2} \right)}{\sin \left( \dfrac{y}{2} \right)}\sin \left( x+\dfrac{N-1}{2}y \right)$ .
By comparing the formulae and expressions we will derive the values of $x,y\text{ }and\text{ N}$ .
We have the first term as $\sin \dfrac{\pi }{n}$ in the given expression which has to be compared with the first term in the formulae which is $\sin \left( x \right)$ which implies $x=\dfrac{\pi }{n}$ .
We have the second term as $\sin \dfrac{3\pi }{n}$ in the given expression which has to be compared with the second term in the formulae which is $\sin \left( x+y \right)$by using the value of $x$ in it. We will get the value of $y$ as $y=\dfrac{2\pi }{n}$ .
We have the last term as $\sin \dfrac{\left( 2n+1 \right)\pi }{n}$ in the given expression which has to be compared with the last term in the formulae which is $\sin \left( x+\left( N-1 \right)y \right)$ by using the value of $x$ and $y$ in it. We will get the value of $N$ as $N=n+1$ .
Therefore we have $x=\dfrac{\pi }{n}$ ,$y=\dfrac{2\pi }{n}$ and $N=n+1$ .
By using these values we will have $\sin \dfrac{\pi }{n}+\sin \dfrac{3\pi }{n}+\sin \dfrac{5\pi }{n}+.............\text{to n terms}=\dfrac{\sin \left( \dfrac{\left( n+1 \right)\dfrac{2\pi }{n}}{2} \right)}{\sin \left( \dfrac{\left( \dfrac{2\pi }{n} \right)}{2} \right)}\sin \left( \dfrac{\pi }{n}+\dfrac{n}{2}\left( \dfrac{2\pi }{n} \right) \right)$
By simplifying this we will have
$\Rightarrow \dfrac{\sin \left( \left( n+1 \right)\dfrac{\pi }{n} \right)}{\sin \left( \dfrac{\pi }{n} \right)}\sin \left( \dfrac{\pi }{n}+\pi \right)$
Since we know that $\sin \left( \pi +x \right)=-\sin x$ we will have
$ \Rightarrow \dfrac{\sin \left( \left( n+1 \right)\dfrac{\pi }{n} \right)}{\sin \left( \dfrac{\pi }{n} \right)}\sin \left( \dfrac{\pi }{n}+\pi \right) $
$ \Rightarrow \dfrac{\sin \left( \left( n+1 \right)\dfrac{\pi }{n} \right)}{\sin \left( \dfrac{\pi }{n} \right)}\left( -\sin \left( \dfrac{\pi }{n} \right) \right) $
$ \Rightarrow - \sin \left( \left( n+1 \right)\dfrac{\pi }{n} \right) $
By further simplifying this by using $\sin \pi n=0$ for any value of $n$ we will have
 $\Rightarrow - \sin \left( \left( n+1 \right)\dfrac{\pi }{n} \right) $
$ \Rightarrow - \sin \left( \left( 1+\dfrac{1}{n} \right)\pi \right) $
$ \Rightarrow 0 $
Therefore we can conclude that $\sin \dfrac{\pi }{n}+\sin \dfrac{3\pi }{n}+\sin \dfrac{5\pi }{n}+.............\text{to n terms}=0$ .
Hence option B is correct.

Note: While answering questions of this type we should be aware of the simply trigonometric conversions like $\sin \left( \pi +x \right)=-\sin x$, $\sin \left( \pi -x \right)=\sin x$ , $\sin \left( 2\pi +x \right)=\sin x$ and $\sin \left( 2\pi -x \right)=-\sin x$ similarly we have formulae for all other trigonometric formulae which can be found in different books or websites. We can make mistakes while using these formulae like in this case we had forgotten to put negative signs. It will not make any difference in this question but it will make a lot of differences in other questions of this type.