Question

The sum of the series $\sum _{\mathrm{n}=1}^{\infty }\sin \left(\frac{\mathrm{n}!\mathrm{\pi }}{720}\right)$ is

• A. $\sin \left(\frac{\mathrm{\pi }}{180}\right)+\sin \left(\frac{\mathrm{\pi }}{360}\right)+\sin \left(\frac{\mathrm{\pi }}{540}\right)$
• B. $\sin \left(\frac{\mathrm{\pi }}{6}\right)+\sin \left(\frac{\mathrm{\pi }}{30}\right)+\sin \left(\frac{\mathrm{\pi }}{120}\right)+\sin \left(\frac{\mathrm{\pi }}{360}\right)$
• C. $\sin \left(\frac{\mathrm{\pi }}{6}\right)+\sin \left(\frac{\mathrm{\pi }}{30}\right)+\sin \left(\frac{\mathrm{\pi }}{120}\right)+\sin \left(\frac{\mathrm{\pi }}{360}\right)+\sin \left(\frac{\mathrm{\pi }}{720}\right)$
• D. $\sin \left(\frac{\mathrm{\pi }}{180}\right)+\sin \left(\frac{\mathrm{\pi }}{360}\right)$

Answer 1

The correct option is C

$\sin \left(\frac{\mathrm{\pi }}{6}\right)+\sin \left(\frac{\mathrm{\pi }}{30}\right)+\sin \left(\frac{\mathrm{\pi }}{120}\right)+\sin \left(\frac{\mathrm{\pi }}{360}\right)+\sin \left(\frac{\mathrm{\pi }}{720}\right)$

Explanation of the correct option.

Given : $\sum _{\mathrm{n}=1}^{\infty }\sin \left(\frac{\mathrm{n}!\mathrm{\pi }}{720}\right)$

$$\begin{gathered} =\sin \left(\frac{\mathrm{\pi }}{720}\right)+\sin \left(\frac{2!\mathrm{\pi }}{720}\right)+\sin \left(\frac{3!\mathrm{\pi }}{720}\right)+\sin \left(\frac{4!\mathrm{\pi }}{720}\right)+\sin \left(\frac{5!\mathrm{\pi }}{720}\right)+\sin \left(\frac{6!\mathrm{\pi }}{720}\right)+\sin \left(\frac{7!\mathrm{\pi }}{720}\right)...................\infty terms \\ =\sin \left(\frac{\mathrm{\pi }}{720}\right)+\sin \left(\frac{\mathrm{\pi }}{360}\right)+\sin \left(\frac{\mathrm{\pi }}{120}\right)+\sin \left(\frac{\mathrm{\pi }}{30}\right)+\sin \left(\frac{\mathrm{\pi }}{6}\right)+\sin \left(\mathrm{\pi }\right)+\sin \left(7\mathrm{\pi }\right)+\sin \left(56\mathrm{\pi }\right)..................\infty terms \end{gathered}$$

We know that, $\sin \left(\mathrm{n\pi }\right)=0,\forall \mathrm{n}\in \mathrm{N}$

Thus in the above expression, the terms after the first five will be equal to $0$.

Therefore,

$\begin{array}{rcl}\sum _{\mathrm{n}=1}^{\infty }\sin \left(\frac{\mathrm{n}!\mathrm{\pi }}{720}\right)& =& \sin \left(\frac{\mathrm{\pi }}{720}\right)+\sin \left(\frac{\mathrm{\pi }}{360}\right)+\sin \left(\frac{\mathrm{\pi }}{120}\right)+\sin \left(\frac{\mathrm{\pi }}{30}\right)+\sin \left(\frac{\mathrm{\pi }}{6}\right)\\ & =& \sin \left(\frac{\mathrm{\pi }}{6}\right)+\sin \left(\frac{\mathrm{\pi }}{30}\right)+\sin \left(\frac{\mathrm{\pi }}{120}\right)+\sin \left(\frac{\mathrm{\pi }}{360}\right)+\sin \left(\frac{\mathrm{\pi }}{720}\right)\end{array}$

Hence, option $\left(C\right)$ is the correct option.