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The sum of the series \[\sum\limits_{n=1}^{\infty }{\sin \left( \dfrac{n!\pi }{720} \right)}\] is
a) \[\sin \left( \dfrac{\pi }{180} \right)+\sin \left( \dfrac{\pi }{360} \right)+\sin \left( \dfrac{\pi }{540} \right)\]
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) \[\sin \left( \dfrac{\pi }{180} \right)+\sin \left( \dfrac{\pi }{360} \right)\]
Answer
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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 .
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 .
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