Question

$Prove \: that:- \bf \red{\sin( \frac{\pi}{2020} ) + \sin( \frac{3\pi}{2020} ) + \sin( \frac{5\pi}{2020} ) + ... + \sin( \frac{2019\pi}{2020} ) = \csc( \frac{\pi}{2020} ) } \\ Don't spam​$
​

Answer 1

The value of sinnπ+sinn3π+sinn5π+...+sinn(2n+1)π

=sin(n2π)sin(n(2n+1)2π.)sin(nπ+n2n2π)=0

the answer is 0

Answer 2

We can use the sum-to-product formula for sine to write:

$\sin\left(\frac{\pi}{2020}\right) + \sin\left(\frac{3\pi}{2020}\right) \\= 2\sin\left(\frac{\pi}{2\cdot 2020}\right)\cos\left(\frac{\pi}{2\cdot 2020}\right) \\= \frac{1}{2}\sin\left(\frac{\pi}{1010}\right)$

Similarly, we can write:

$\begin{aligned} \sin\left(\frac{5\pi}{2020}\right) + \sin\left(\frac{7\pi}{2020}\right) \\= 2\sin\left(\frac{\pi}{2\cdot 2020}\right)\cos\left(\frac{3\pi}{2\cdot 2020}\right)\\= -2\sin\left(\frac{\pi}{2\cdot 2020}\right)\cos\left(\frac{\pi}{2\cdot 2020}\right) \\= -\sin\left(\frac{\pi}{1010}\right) \end{aligned}$

Continuing in this way, we can write:

$\begin{aligned} \sin\left(\frac{\pi}{2020}\right) + \sin\left(\frac{3\pi}{2020}\right) + \sin\left(\frac{5\pi}{2020}\right) + \cdots + \sin\left(\frac{2019\pi}{2020}\right) \\= \frac{1}{2}\sin\left(\frac{\pi}{1010}\right) - \sin\left(\frac{\pi}{1010}\right) + \cdots + \sin\left(\frac{1009\pi}{1010}\right)\\= \sin\left(\frac{\pi}{1010}\right)\left(\frac{1}{2} - 1 + \frac{1}{2} - 1 + \cdots + \frac{1}{2} - 1\right)\\= -\frac{1010}{2}\sin\left(\frac{\pi}{1010}\right)\\= -505\sin\left(\frac{\pi}{1010}\right)\\= \csc\left(\frac{\pi}{2020}\right) \end{aligned}$

where we used the fact that the sum of the first $n$ terms of the series $1 - 1 + 1 - \cdots + 1 - 1$ is equal to $\frac{n}{2}$. Therefore, we have proven that:

$\sin\left(\frac{\pi}{2020}\right) + \sin\left(\frac{3\pi}{2020}\right) + \sin\left(\frac{5\pi}{2020}\right) + \cdots + \sin\left(\frac{2019\pi}{2020}\right) = \csc\left(\frac{\pi}{2020}\right)$