Question

Given 15 theta =pi
This is from trigonometry​

Answer 1

$\large\underline{\bold{Given \:Question - }}$

Prove that

$\rm :\longmapsto\:cos\theta cos2\theta cos3\theta cos4\theta cos5\theta cos6\theta cos7\theta = \dfrac{1}{ {2}^{7} }$

when

$\rm :\longmapsto\:15\theta = \pi$

$\begin{gathered}\large{\sf{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\boxed{ \bf{ \: cos(\pi - \theta ) = - cos\theta }}$

$\boxed{ \bf{ \: sin(\pi - \theta ) = sin\theta }}$

$\boxed{ \bf{ \: sin(\pi + \theta ) = - \: sin\theta }}$

$\boxed{ \bf{ \: cosxcos {2}^{2}xcos {2}^{3}x - - - - cos {2}^{n}x = \frac{sin {2}^{n + 1} x}{ {2}^{n + 1} sinx} }}$

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

Given that

$\rm :\longmapsto\: 15\theta \: = \: \pi$

Now, Consider,

$\rm :\longmapsto\:cos\theta cos2\theta cos3\theta cos4\theta cos5\theta cos6\theta cos7\theta$

can be rewritten as

$\rm \: = \:cos\theta cos2\theta cos3\theta cos4\theta cos5\theta cos6\theta cos(15\theta - 8\theta )$

$\rm \: = \:cos\theta cos2\theta cos3\theta cos4\theta cos5\theta cos6\theta cos(\pi - 8\theta )$

$\rm \: = - \:cos\theta cos2\theta cos3\theta cos4\theta cos5\theta cos6\theta cos8\theta$

can be further rearrange as

$\rm \: = \: - \: [cos\theta cos2\theta cos4\theta cos8\theta ] \: [cos3\theta cos6\theta ] \: cos5\theta$

$\rm \: = \: - [cos\theta cos2\theta cos {2}^{2}\theta cos {2}^{3}\theta ] \: [cos3\theta cos2(3\theta )] \: cos5\theta$

$\rm \: = \: - \: \bigg[\dfrac{sin {2}^{3 + 1} \theta }{ {2}^{3 + 1}sin\theta } \bigg] \: \bigg[\dfrac{sin {2}^{1 + 1}3\theta }{ {2}^{1 + 1} sin3\theta } \bigg] \: cos5\theta$

$\rm \: = \: - \: \bigg[\dfrac{sin {2}^{4} \theta }{ {2}^{4}sin\theta } \bigg] \: \bigg[\dfrac{sin {2}^{2}(3\theta ) }{ {2}^{2} sin3\theta } \bigg] \: cos5\theta$

$\rm \: = \: - \: \bigg[\dfrac{sin 16 \theta }{ {2}^{4}sin\theta } \bigg] \: \bigg[\dfrac{sin12\theta}{ {2}^{2} sin3\theta } \bigg] \: cos5\theta$

$\rm \: = \: - \: \bigg[\dfrac{sin( 15 \theta + \theta ) }{ {2}^{4}sin\theta } \bigg] \: \bigg[\dfrac{sin(15\theta - 3\theta )}{ {2}^{2} sin3\theta } \bigg] \: cos5\theta$

$\rm \: = \: - \: \bigg[\dfrac{sin( \pi+ \theta ) }{ {2}^{4}sin\theta } \bigg] \: \bigg[\dfrac{sin(\pi - 3\theta )}{ {2}^{2} sin3\theta } \bigg] \: cos5\theta$

$\rm \: = \: - \: \bigg[ - \dfrac{sin\theta }{ {2}^{4}sin\theta } \bigg] \: \bigg[\dfrac{sin3\theta }{ {2}^{2} sin3\theta } \bigg] \: cos5\theta$

$\rm \: = \:\dfrac{1}{ {2}^{6} } \times cos\bigg[5 \times \dfrac{\pi}{15} \bigg]$

$\rm \: = \:\dfrac{1}{ {2}^{6} } \times cos\bigg[\dfrac{\pi}{3} \bigg]$

$\rm \: = \:\dfrac{1}{ {2}^{6} } \times \dfrac{1}{2}$

$\rm \: = \:\dfrac{1}{ {2}^{7} }$

Hence,

$\rm :\longmapsto\:cos\theta cos2\theta cos3\theta cos4\theta cos5\theta cos6\theta cos7\theta = \dfrac{1}{ {2}^{7} }$