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# The sum of the series $\sum\limits_{n=1}^{\infty }{\sin \left( \dfrac{n!\pi }{720} \right)}$ isa) $\sin \left( \dfrac{\pi }{180} \right)+\sin \left( \dfrac{\pi }{360} \right)+\sin \left( \dfrac{\pi }{540} \right)$ b) $\sin \left( \dfrac{\pi }{6} \right)+\sin \left( \dfrac{\pi }{30} \right)+\sin \left( \dfrac{\pi }{120} \right)+\sin \left( \dfrac{\pi }{360} \right)$c) $\sin \left( \dfrac{\pi }{6} \right)+\sin \left( \dfrac{\pi }{30} \right)+\sin \left( \dfrac{\pi }{120} \right)+\sin \left( \dfrac{\pi }{360} \right)+\sin \left( \dfrac{\pi }{720} \right)$ d) $\sin \left( \dfrac{\pi }{180} \right)+\sin \left( \dfrac{\pi }{360} \right)$

Last updated date: 20th Mar 2023
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Hint: The value of $sine$ function for any argument that is an integral multiple of $\pi$ is equal to $0$ , i.e. $\sin (n\pi )=0$ where $n=1,2,3,4...$ and so on.

Before we proceed with the solution , we must know some properties of the $sine$ function .
a) The value of $\sin (\pi )$ is equal to $0$.
b) $\sin (\pi +\theta )=-\sin \theta$
Now , to find the value of $\sin (2\pi )$ , we will substitute $\theta =\pi$ in property $(b)$.
So , we get $\sin (2\pi )=\sin \left( \pi +\pi \right)=-\sin \pi =0$ .
Similarly , $\sin \left( 3\pi \right)=\sin \left( \pi +2\pi \right)=-\sin 2\pi =0$. So , following the pattern , we can conclude that the value of $sine$ function for any argument that is an integral multiple of $\pi$ is equal to $0$, i.e.
$\sin (n\pi )=0$ where $n=1,2,3,4...$and so on.
Now , coming to the question , we need to find the value of the sum $\sum\limits_{n=1}^{\infty }{\sin \left( \dfrac{n!\pi }{720} \right)}$.
First , we will open the summation sign and write it as a sum of a series.
So , we can write $\sum\limits_{n=1}^{\infty }{\sin \left( \dfrac{n!\pi }{720} \right)}$ as $\sum\limits_{n=1}^{\infty }{\sin \left( \dfrac{n!\pi }{720} \right)}=\sin \left( \dfrac{\pi }{720} \right)+\sin \left( \dfrac{2\pi }{720} \right)+\sin \left( \dfrac{6\pi }{720} \right)+\sin \left( \dfrac{24\pi }{720} \right)+\sin \left( \dfrac{120\pi }{720} \right)....$
Now , when $n=6$ , we get $\sin \left( \dfrac{n!\pi }{720} \right)=\sin \left( \dfrac{6!\pi }{720} \right)=\sin \left( \dfrac{720\pi }{720} \right)=\sin \pi$.
But , from the property$(a)$ of $sine$ function , we know the value of $\sin (\pi )$ is equal to $0$.
So , the value of $\sin \left( \dfrac{n!\pi }{720} \right)$, when $n=6$ is equal to $0$.
Now , when $n=7$, we get $\sin \left( \dfrac{n!\pi }{720} \right)=\sin \left( \dfrac{7!\pi }{720} \right)$ .
We know , $n!=n\times (n-1)!$.
So , $7!=7\times 6!=7\times 720$.
So , $\sin \left( \dfrac{7!\pi }{720} \right)=\sin \left( \dfrac{7\times 720\pi }{720} \right)=\sin 7\pi$.
Now , we know , the value of sine function for any argument that is an integral multiple of $\pi$ is equal to $0$.
So , $\sin 7\pi =0$.
Similarly , for $n=8,9,....$and so on , we can see that the value of $\sin \left( \dfrac{n!\pi }{720} \right)$ is equal to $0$.
So , we can write the sum as $\sum\limits_{n=1}^{\infty }{\sin \left( \dfrac{n!\pi }{720} \right)}=\sin \left( \dfrac{\pi }{720} \right)+\sin \left( \dfrac{2\pi }{720} \right)+\sin \left( \dfrac{6\pi }{720} \right)+\sin \left( \dfrac{24\pi }{720} \right)+\sin \left( \dfrac{120\pi }{720} \right)+0+0+0....$
So , $\sum\limits_{n=1}^{\infty }{\sin \left( \dfrac{n!\pi }{720} \right)}=\sin \left( \dfrac{\pi }{720} \right)+\sin \left( \dfrac{2\pi }{720} \right)+\sin \left( \dfrac{6\pi }{720} \right)+\sin \left( \dfrac{24\pi }{720} \right)+\sin \left( \dfrac{120\pi }{720} \right)$
Or , $\sum\limits_{n=1}^{\infty }{\sin \left( \dfrac{n!\pi }{720} \right)}=\sin \left( \dfrac{\pi }{720} \right)+\sin \left( \dfrac{\pi }{360} \right)+\sin \left( \dfrac{\pi }{120} \right)+\sin \left( \dfrac{\pi }{30} \right)+\sin \left( \dfrac{\pi }{6} \right)$
Or , $\sum\limits_{n=1}^{\infty }{\sin \left( \dfrac{n!\pi }{720} \right)}=\sin \left( \dfrac{\pi }{6} \right)+\sin \left( \dfrac{\pi }{30} \right)+\sin \left( \dfrac{\pi }{120} \right)+\sin \left( \dfrac{\pi }{360} \right)+\sin \left( \dfrac{\pi }{720} \right)$
Hence , the value of the sum of the series $\sum\limits_{n=1}^{\infty }{\sin \left( \dfrac{n!\pi }{720} \right)}$ is equal to $\sin \left( \dfrac{\pi }{6} \right)+\sin \left( \dfrac{\pi }{30} \right)+\sin \left( \dfrac{\pi }{120} \right)+\sin \left( \dfrac{\pi }{360} \right)+\sin \left( \dfrac{\pi }{720} \right)$.
Therefore , option (c) is the correct answer.

Note: Students generally get confused between the values of $\sin \pi$ and $\cos \pi$. $\sin \pi =0$ and $\cos \pi =-1$ . Confusion between the values should be avoided as it can lead to wrong answers .