Question

$\tiny\sf \pink{\sin( \frac{\pi}{2020} ) + \sin( \frac{3\pi}{2020} ) + \sin( \frac{5\pi}{2020} ) + ... + \sin( \frac{2019\pi}{2020} ) = \csc( \frac{\pi}{2020} ) }$​

Answer 1

$\large\underline{\sf{Solution-}}$

Consider,

$\rm :\longmapsto\:\sf {\sin( \dfrac{\pi}{2020} ) + \sin( \dfrac{3\pi}{2020} ) + \sin( \dfrac{5\pi}{2020} ) + ... + \sin( \dfrac{2019\pi}{2020} )}$

Now,

$\rm :\longmapsto\:\dfrac{\pi}{2020}, \: \dfrac{3\pi}{2020},\dfrac{5\pi}{2020}, - - \dfrac{2019\pi}{2020} \: forms \: an \: AP \: series$

having

$\rm :\longmapsto\:a = \dfrac{\pi}{2020}$

$\rm :\longmapsto\:d = \dfrac{3\pi}{2020} - \dfrac{\pi}{2020} = \dfrac{2\pi}{2020} = \dfrac{\pi}{1010}$

We know, nth term of AP is given by

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• aₙ is the nᵗʰ term.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.

Tʜᴜs,

$\rm :\longmapsto\:\dfrac{2019\pi}{2020} = \dfrac{\pi}{2020} + (n - 1)\dfrac{2\pi}{2020}$

$\rm :\longmapsto\:2019 = 1 + 2(n - 1)$

$\rm :\longmapsto\:2019 = 1 + 2n - 2$

$\rm :\longmapsto\:2019 = 2n - 1$

$\rm :\longmapsto\:2020= 2n$

$\rm\implies \:n = 1010$

So, it means, above expression can be rewritten as

$\rm :\longmapsto\:\sf {\sin( \dfrac{\pi}{2020} ) + \sin(\dfrac{\pi}{2020} + \dfrac{2\pi}{2020} ) + \sin(\dfrac{\pi}{2020} + 2\dfrac{2\pi}{2020} ) + ... + \sin(\dfrac{\pi}{2020} + \dfrac{2018\pi}{2020} )}$

$\rm =  \sf {\sin( \dfrac{\pi}{2020} ) + \sin(\dfrac{\pi}{2020} + \dfrac{2\pi}{2020} ) + \sin(\dfrac{\pi}{2020} + 2\dfrac{2\pi}{2020} ) + ... + \sin(\dfrac{\pi}{2020} + \dfrac{1009 \times 2\pi}{2020} )}$

$\rm =  \sf {\sin( \dfrac{\pi}{2020} ) + \sin(\dfrac{\pi}{2020} + \dfrac{2\pi}{2020} ) + \sin(\dfrac{\pi}{2020} + 2\dfrac{2\pi}{2020} ) + ... + \sin(\dfrac{\pi}{2020} +(1010 - 1) \dfrac{2\pi}{2020} )}$

We know,

$\boxed{\sf{sina + sin (a + d) + sin (a + 2d) + - - + sin (a + (n - 1)d) = \dfrac{sin \bigg(a + \dfrac{n - 1}{2}d\bigg)sin \bigg( \dfrac{nd}{2} \bigg)}{sin \bigg( \dfrac{d}{2} \bigg)} }}$

So, here we have

$\rm :\longmapsto\:a = \dfrac{\pi}{2020}$

$\rm :\longmapsto\:d = \dfrac{\pi}{1010}$

$\rm :\longmapsto\:n = 1010$

So, on substituting the values, we get

$\rm \:  =  \: \dfrac{sin \bigg(\dfrac{\pi}{2020} + \dfrac{1010 - 1}{2} \times \dfrac{\pi}{1010} \bigg)sin \bigg( 1010 \times \dfrac{\pi}{2020} \bigg)}{sin \dfrac{\pi}{2020}}$

$\rm \:  =  \: \dfrac{sin \bigg(\dfrac{\pi}{2020} + \dfrac{1009\pi}{2020} \bigg)sin \bigg( \dfrac{\pi}{2} \bigg)}{sin \dfrac{\pi}{2020}}$

$\rm \:  =  \: \dfrac{sin \bigg( \dfrac{\pi + 1009\pi}{2020} \bigg)}{sin \dfrac{\pi}{2020}}$

$\rm \:  =  \: \dfrac{sin \bigg( \dfrac{1010\pi}{2020} \bigg)}{sin \dfrac{\pi}{2020}}$

$\rm \:  =  \: \dfrac{sin \bigg( \dfrac{\pi}{2} \bigg)}{sin \dfrac{\pi}{2020}}$

$\rm \:  =  \: \dfrac{1}{sin \dfrac{\pi}{2020}}$

$\rm \:  =  \: cosec\dfrac{\pi}{2020}$

Hence, Proved