Question

Find sinA+ sin(π+A) + sin(2π+A) + sin(3π+A)+.......................... upto 2021 term?

प्लीज टेल मी द आंसर ​

Answer 1

Answer:

sinA-SinA+sinA-sinA+......sinA-sinA(2020 terms)+sinA=sinA

Answer 2

Given to find,

$\displaystyle\longrightarrow S=\underbrace{\sin A+\sin(\pi+A)+\sin(2\pi+A)+\,\dots\,+\sin(2020\pi+A)}_{2021\ terms}$

Or we can say that,

$\displaystyle\longrightarrow S=\sum_{r=0}^{2020}\sin(r\pi+A)$

We're going to divide this sum into two sums - one contains the terms in which value of $r$ is odd, and the other contains those in which value of $r$ is even.

The sum of terms in this sum $S,$ where $r$ has odd value, is,

$\displaystyle\longrightarrow S_o=\sin(\pi+A)+\sin(3\pi+A)+\sin(5\pi+A)+\,\dots\,+\sin(2019\pi+A)$

$\displaystyle\longrightarrow S_o=\sum_{r=1}^{1010}\sin((2r-1)\pi+A)$

The sum of terms in this sum $S,$ where $r$ has even value, is,

$\displaystyle\longrightarrow S_e=\sin A+\sin(2\pi+A)+\sin(4\pi+A)+\,\dots\,+\sin(2020\pi+A)$

$\displaystyle\longrightarrow S_e=\sum_{r=0}^{1010}\sin(2r\pi+A)$

Hence,

$\displaystyle\longrightarrow S=S_o+S_e$

$\displaystyle\longrightarrow S=\sum_{r=1}^{1010}\sin((2r-1)\pi+A)+\sum_{r=0}^{1010}\sin(2r\pi+A)$

$\displaystyle\longrightarrow S=\sum_{r=1}^{1010}\sin(2\pi r-\pi+A)+\sum_{r=0}^{1010}\sin(2\pi r+A)$

$\displaystyle\longrightarrow S=\sum_{r=1}^{1010}\sin(2\pi r-(\pi-A))+\sum_{r=0}^{1010}\sin(2\pi r+A)$

Since $\sin(2\pi n+x)=\sin x$ where $n\in\mathbb{Z},$

$\displaystyle\longrightarrow S=\sum_{r=1}^{1010}\sin(-(\pi-A))+\sum_{r=0}^{1010}\sin A$

Since $\sin(-x)=-\sin x,$

$\displaystyle\longrightarrow S=\sum_{r=1}^{1010}-\sin(\pi-A)+\sum_{r=0}^{1010}\sin A$

$\displaystyle\longrightarrow S=-\sum_{r=1}^{1010}\sin(\pi-A)+\sum_{r=0}^{1010}\sin A$

Since $\sin(\pi-x)=\sin x,$

$\displaystyle\longrightarrow S=-\sum_{r=1}^{1010}\sin A+\sum_{r=0}^{1010}\sin A$

Since $\displaystyle\sum_{r=a}^bc=(b-a+1)c,$

$\displaystyle\longrightarrow S=-(1010-1+1)\sin A+(1010-0+1)\sin A$

$\displaystyle\longrightarrow S=-1010\sin A+1011\sin A$

$\displaystyle\longrightarrow S=(-1010+1011)\sin A$

$\displaystyle\longrightarrow\underline{\underline{S=\sin A}}$

Hence sin A is the answer.