Question

Prove the following:-

$\bigstar \boxed{\displaystyle \sf\sum\limits_{r = 0}^{n} { \sin}^{2} \bigg( \frac{2 \pi}{n} \bigg) = \frac{n}{2} }$

Class - 11th

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Answer 1

Appropriate Question :-

Prove the following:-

$\bigstar \boxed{\displaystyle \sf\sum\limits_{r = 0}^{n} { \sin}^{2} \bigg( \frac{2r \pi}{n} \bigg) = \frac{n}{2} }$

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

Given that,

$\rm :\longmapsto\:{\displaystyle \sf\sum\limits_{r = 0}^{n} { \sin}^{2} \bigg( \frac{2r \pi}{n} \bigg) }$

Let we assume that,

$\red{\rm :\longmapsto\:\dfrac{2\pi}{n} = x}$

So, given can be rewritten as

$\rm :\longmapsto\:{\displaystyle \sf\sum\limits_{r = 0}^{n} { \sin}^{2} \bigg( rx \bigg) }$

$\rm \:  =  \:  \: \:{\displaystyle \: \frac{1 }{2} \: \sf\sum\limits_{r = 0}^{n} { 2\sin}^{2} \bigg( rx \bigg) }$

$\rm \:  =  \:  \: \:{\displaystyle \: \frac{1 }{2} \: \sf\sum\limits_{r = 0}^{n}(1 - cos2rx) }$

$\rm \:  =  \:  \: \:{\displaystyle \: \frac{1 }{2} \: \sf\sum\limits_{r = 0}^{n} 1} - \:{\displaystyle \: \frac{1 }{2} \: \sf\sum\limits_{r = 0}^{n}(cos2rx) }$

$\rm \:  =  \dfrac{1}{2}(n + 1) - \dfrac{1}{2}(cos0 + cos2x + cos4x + - - cos2nx)$

$\rm \:  =  \dfrac{1}{2}(n + 1) - \dfrac{1}{2}(1 + cos2x + cos4x + - - cos2nx)$

$\rm \:  =  \dfrac{1}{2}(n + 1) - \dfrac{1}{2} - \dfrac{1}{2}(cos2x + cos4x + - - cos2nx)$

$\rm \:  =  \dfrac{1}{2}(n + 1 - 1) - \dfrac{1}{2}(cos2x + cos4x + - - cos2nx)$

$\rm \:  =  \dfrac{1}{2}(n) - \dfrac{1}{2}(cos2x + cos4x + - - cos2nx)$

$\rm \:  =  \dfrac{1}{2}(n) - \dfrac{1}{2}(cos2x + cos(2x + 2x) + - - cos(2x + (n - 1)2x)$

We know,

$\green{\rm :\longmapsto\:cosa + cos(a + d) + cos(a + 2d) + - - + cos(a + (n - 1)d} \\ \green{\rm \: = \: \dfrac{sin\dfrac{nd}{2} \: cos\dfrac{1}{2} (2a + (n - 1)d)}{sin\dfrac{d}{2} } }$

So, using this identity, we get

$\rm \:  =  \:  \: \dfrac{n}{2} - \dfrac{1}{2}\bigg(\dfrac{sinnx \: cos\dfrac{1}{2}(4x + (n - 1)2x)}{sinx} \bigg)$

$\rm \:  =  \:  \: \dfrac{n}{2} - \dfrac{1}{2}\bigg(\dfrac{sinnx \: cos(2x + (n - 1)x)}{sinx} \bigg)$

$\rm \:  =  \:  \: \dfrac{n}{2} - \dfrac{1}{2}\bigg(\dfrac{sinnx \: cos(2x + nx - x)}{sinx} \bigg)$

$\rm \:  =  \:  \: \dfrac{n}{2} - \dfrac{1}{2}\bigg(\dfrac{sinnx \: cos(x + nx)}{sinx} \bigg)$

$\rm \:  =  \:  \: \dfrac{n}{2} - \dfrac{1}{2}\bigg(\dfrac{sinnx \: cos(1 + n)x}{sinx} \bigg)$

Now, Substitute the value of x, we get

$\rm \:  =  \:  \: \dfrac{n}{2} - \dfrac{1}{2}\bigg(\dfrac{sin2\pi \: cos(1 + n)\dfrac{2\pi}{n} }{sin\dfrac{2\pi}{n} } \bigg)$

We know,

$\boxed{ \rm{ sin2\pi = 0}}$

So, above can be rewritten as

$\rm \:  =  \:  \: \dfrac{n}{2} - 0$

$\rm \:  =  \:  \: \dfrac{n}{2}$

Hence,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\displaystyle \bf\sum\limits_{r = 0}^{n} { \sin}^{2} \bigg( \frac{2r \pi}{n} \bigg) = \frac{n}{2} }$

Answer 2

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