Question

Please help solve this question.

Answer 1

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

$\rm \: P(n) : sin \theta + sin2\theta + sin3\theta + - - - + sinn\theta = \dfrac{sin\dfrac{(n + 1)\theta }{2} sin\dfrac{n\theta }{2}}{sin\dfrac{\theta }{2}}$

Step :- 1 For n = 1, we have

$\rm \: P(1) : sin \theta = \dfrac{sin\dfrac{(1 + 1)\theta }{2} sin\dfrac{\theta }{2}}{sin\dfrac{\theta }{2}}$

$\rm :\implies\:sin\theta = sin\theta$

$\rm :\implies\:P(n) \: is \: true \: for \: n = 1$

Step :- 2 Let assume that statement P(n) is true for n = k, where k is some natural number.

$\rm \: P(k) : sin \theta + sin2\theta + sin3\theta + - - - + sink\theta = \dfrac{sin\dfrac{(k + 1)\theta }{2} sin\dfrac{k\theta }{2}}{sin\dfrac{\theta }{2}}$

Step : - 3 We have to prove that statement P(n) is true for n = k + 1.

$\rm \: P(k + 1) : sin \theta + sin2\theta + sin3\theta + - - - + sin(k + 1)\theta = \dfrac{sin\dfrac{(k + 2)\theta }{2} sin\dfrac{(k + 1)\theta }{2}}{sin\dfrac{\theta }{2}}$

Consider,

$\rm :\longmapsto\:sin \theta + sin2\theta + sin3\theta + - - - + sin(k + 1)\theta$

$\rm \:  =  \:  \: \dfrac{sin\dfrac{(k + 1)\theta }{2} sin\dfrac{k\theta }{2}}{sin\dfrac{\theta }{2}} + sin(k + 1)\theta$

$\rm \:  =  \:  \: \dfrac{sin\dfrac{(k + 1)\theta }{2} sin\dfrac{k\theta }{2}}{sin\dfrac{\theta }{2}} + 2sin\dfrac{(k+ 1)\theta }{2}cos\dfrac{(k+ 1)\theta }{2}$

$\rm \:  =  \:  \: sin\dfrac{(k+ 1)\theta }{2}\bigg(\dfrac{sin\dfrac{k\theta }{2}}{sin \dfrac{\theta }{2} } + 2cos\dfrac{(k+ 1)\theta }{2} \bigg)$

$\rm \:  =  \:  \: sin\dfrac{(k+ 1)\theta }{2}\bigg(\dfrac{sin\dfrac{k\theta }{2} + + 2cos\dfrac{(k+ 1)\theta }{2}sin \dfrac{\theta }{2} }{sin \dfrac{\theta }{2} } \bigg)$

$\rm \:  =  \:  \: sin\dfrac{(k+ 1)\theta }{2}\bigg(\dfrac{sin\dfrac{k\theta }{2} + sin\bigg(\dfrac{(k + 1)\theta }{2} + \dfrac{\theta }{2} \bigg) +sin\bigg(\dfrac{(k + 1)\theta }{2} - \dfrac{\theta }{2} \bigg) }{sin \dfrac{\theta }{2} } \bigg)$

$\rm \:  =  \:  \: sin\dfrac{(k+ 1)\theta }{2}\bigg(\dfrac{sin\dfrac{k\theta }{2} + sin\bigg(\dfrac{(k + 1)\theta }{2} + \dfrac{\theta }{2} \bigg) - sin\bigg( \dfrac{k\theta }{2} \bigg) }{sin \dfrac{\theta }{2} } \bigg)$

$\rm \:  =  \:  \: \dfrac{sin\dfrac{(k + 2)\theta }{2} sin\dfrac{(k + 1)\theta }{2}}{sin\dfrac{\theta }{2}}$

Hence,

By the Process of Principal of Mathematical Induction,

$\rm \: P(n) : sin \theta + sin2\theta + sin3\theta + - - - + sinn\theta = \dfrac{sin\dfrac{(n + 1)\theta }{2} sin\dfrac{n\theta }{2}}{sin\dfrac{\theta }{2}}$

Formula Used :-

$\boxed{ \rm{ sin2x = 2sinxcosx}}$

$\boxed{ \rm{ 2cosxsiny = sin(x + y) - sin(x - y)}}$