Question

Prove that:-

$\bf \red{\sin( \frac{\pi}{2020} ) + \sin( \frac{3\pi}{2020} ) + \sin( \frac{5\pi}{2020} ) + ... + \sin( \frac{2019\pi}{2020} ) = \csc( \frac{\pi}{2020} ) }$

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Answer 1

Consider the LHS.

$\small\longrightarrow a=\sin\left(\dfrac{\pi}{2020}\right)+\sin\left(\dfrac{3\pi}{2020}\right)+\sin\left(\dfrac{5\pi}{2020}\right)+\,\dots\,+\sin\left(\dfrac{2019\pi}{2020}\right)\quad\dots(i)$

Or,

$\displaystyle\small\longrightarrow a=\sum_{r=1}^{1010}\sin\left(\dfrac{(2r-1)\pi}{2020}\right)$

Multiplying both sides by $\small2\sin\left(\dfrac{\pi}{2020}\right),$

$\displaystyle\small\longrightarrow 2a\sin\left(\dfrac{\pi}{2020}\right)=2\sin\left(\dfrac{\pi}{2020}\right)\sum_{r=1}^{1010}\sin\left(\dfrac{(2r-1)\pi}{2020}\right)$

$\displaystyle\small\longrightarrow 2a\sin\left(\dfrac{\pi}{2020}\right)=\sum_{r=1}^{1010}2\sin\left(\dfrac{(2r-1)\pi}{2020}\right)\sin\left(\dfrac{\pi}{2020}\right)\quad\dots(1)$

We have, by product-to-sum identity,

$\small\longrightarrow2\sin A\sin B=\cos(A-B)-\cos(A+B)$

Taking $\smallA=\dfrac{(2r-1)\pi}{2020}$ and $\smallB=\dfrac{\pi}{2020},$

$\small\longrightarrow2\sin\left(\dfrac{(2r-1)\pi}{2020}\right)\sin\left(\dfrac{\pi}{2020}\right)=\cos\left(\dfrac{(r-1)\pi}{1010}\right)-\cos\left(\dfrac{r\pi}{1010}\right)$

Then (1) becomes,

$\displaystyle\small\longrightarrow2a\sin\left(\dfrac{\pi}{2020}\right)=\sum_{r=1}^{1010}\left[\cos\left(\dfrac{(r-1)\pi}{1010}\right)-\cos\left(\dfrac{r\pi}{1010}\right)\right]$

$\displaystyle\small\longrightarrow2a\sin\left(\dfrac{\pi}{2020}\right)=\sum_{r=1}^{1010}\cos\left(\dfrac{(r-1)\pi}{1010}\right)-\sum_{r=1}^{1010}\cos\left(\dfrac{r\pi}{1010}\right)\quad\dots(2)$

We have,

• $\displaystyle\small\sum_{r=p}^qf(r)=\sum_{r=p-t}^{q-t}f(r+t)$

Taking $\smallf(r)=\cos\left(\dfrac{(r-1)\pi}{1010}\right),\ p=1,\ q=1010,\ t=1,$ the first sum in RHS can be written as,

• $\displaystyle\small\sum_{r=1}^{1010}\cos\left(\dfrac{(r-1)\pi}{1010}\right)=\sum_{r=0}^{1009}\cos\left(\dfrac{r\pi}{1010}\right)$

Thus (2) becomes,

$\displaystyle\small\longrightarrow2a\sin\left(\dfrac{\pi}{2020}\right)=\sum_{r=0}^{1009}\cos\left(\dfrac{r\pi}{1010}\right)-\sum_{r=1}^{1010}\cos\left(\dfrac{r\pi}{1010}\right)\quad\dots(3)$

Take,

• $\displaystyle\small\sum_{r=0}^{1009}\cos\left(\dfrac{r\pi}{1010}\right)=\cos\left(\dfrac{(0)\pi}{1010}\right)+\sum_{r=1}^{1009}\cos\left(\dfrac{r\pi}{1010}\right)=1+\sum_{r=1}^{1009}\cos\left(\dfrac{r\pi}{1010}\right)$

[i.e. first term is taken out of the sum]

• $\displaystyle\small\sum_{r=1}^{1010}\cos\left(\dfrac{r\pi}{1010}\right)=\sum_{r=1}^{1009}\cos\left(\dfrac{r\pi}{1010}\right)+\cos\left(\dfrac{1010\pi}{1010}\right)=\sum_{r=1}^{1009}\cos\left(\dfrac{r\pi}{1010}\right)-1$

[i.e. last term is taken out of the sum]

Then (3) becomes,

$\displaystyle\small\longrightarrow2a\sin\left(\dfrac{\pi}{2020}\right)=\left[1+\sum_{r=1}^{1009}\cos\left(\dfrac{r\pi}{1010}\right)\right]-\left[\sum_{r=1}^{1009}\cos\left(\dfrac{r\pi}{1010}\right)-1\right]$

$\displaystyle\small\longrightarrow2a\sin\left(\dfrac{\pi}{2020}\right)=1+\sum_{r=1}^{1009}\cos\left(\dfrac{r\pi}{1010}\right)-\sum_{r=1}^{1009}\cos\left(\dfrac{r\pi}{1010}\right)+1$

$\displaystyle\small\longrightarrow2a\sin\left(\dfrac{\pi}{2020}\right)=2$

$\displaystyle\small\text{ \longrightarrow a=\dfrac{1}{\sin\left(\dfrac{\pi}{2020}\right)} }$

$\displaystyle\small\longrightarrow a=\csc\left(\dfrac{\pi}{2020}\right)\quad\dots(ii)$

From (i) and (ii),

$\small\text{ \longrightarrow\underline{\underline{\sin\left(\dfrac{\pi}{2020}\right)+\sin\left(\dfrac{3\pi}{2020}\right)+\sin\left(\dfrac{5\pi}{2020}\right)+\,\dots\,+\sin\left(\dfrac{2019\pi}{2020}\right)=\csc\left(\dfrac{\pi}{2020}\right)}} }$

Hence Proved!

Answer 2

$\huge\bold \green{Thank \: You}$