Question

Answer this please in paper​

Answer 1

Given Question :-

Prove that

$\sf \: cos\bigg[\dfrac{2\pi}{7} \bigg]cos\bigg[\dfrac{4\pi}{7} \bigg]cos\bigg[\dfrac{8\pi}{7} \bigg] = \dfrac{1}{8}$

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

Consider LHS

$\rm :\longmapsto\: \sf \: cos\bigg[\dfrac{2\pi}{7} \bigg]cos\bigg[\dfrac{4\pi}{7} \bigg]cos\bigg[\dfrac{8\pi}{7} \bigg]$

can be rewritten as

$\rm \:  =  \:\: \sf \: cos\bigg[\dfrac{2\pi}{7} \bigg]cos\bigg[\dfrac{4\pi}{7} \bigg]cos\bigg[\pi + \dfrac{\pi}{7} \bigg]$

We know,

$\boxed{ \rm \:cos(\pi + x) = - cosx}$

So, using this

$\rm \:  =  \:\: - \sf \: cos\bigg[\dfrac{2\pi}{7} \bigg]cos\bigg[\dfrac{4\pi}{7} \bigg]cos\bigg[\dfrac{\pi}{7} \bigg]$

can be re-arranged as

$\rm \:  =  \:\: - \sf \: cos\bigg[\dfrac{\pi}{7} \bigg]cos\bigg[\dfrac{2\pi}{7} \bigg]cos\bigg[\dfrac{4\pi}{7} \bigg]$

$\rm \:  =  \:\: - \sf \: cos\bigg[\dfrac{\pi}{7} \bigg]cos\bigg[\dfrac{2\pi}{7} \bigg]cos\bigg[\dfrac{ {2}^{2} \pi}{7} \bigg]$

We know,

$\boxed{ \rm \:cosx \: cos2x \: cos {2}^{2}x - - - cos {2}^{n}x = \frac{sin {2}^{n + 1}x }{ {2}^{n + 1}sinx}}$

So, using this,

$\rm \:  = \:   - \:\dfrac{sin {2}^{2 + 1}\dfrac{\pi}{7} }{ {2}^{2 + 1}sin\dfrac{\pi}{7} }$

$\rm \:  = \:   - \:\dfrac{sin {2}^{3}\dfrac{\pi}{7} }{ {2}^{3}sin\dfrac{\pi}{7} }$

$\rm \:  = \:   - \:\dfrac{sin \dfrac{8\pi}{7} }{ 8sin\dfrac{\pi}{7} }$

$\rm \:  = \:   - \:\dfrac{sin \bigg[\pi + \dfrac{\pi}{7}\bigg]}{ 8sin\dfrac{\pi}{7} }$

We know,

$\boxed{ \rm \:sin(\pi + x) = - \: sinx}$

So, using this

$\rm \:  = \:   - ( -) \:\dfrac{sin \bigg[ \dfrac{\pi}{7}\bigg]}{ 8sin\dfrac{\pi}{7} }$

$\rm \:  =  \:\dfrac{1}{8}$

Hence,

$\sf \:\bf\implies \: \boxed{ \bf{cos\bigg[\dfrac{2\pi}{7} \bigg]cos\bigg[\dfrac{4\pi}{7} \bigg]cos\bigg[\dfrac{8\pi}{7} \bigg] = \dfrac{1}{8} }}$

Additional Information :-

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

$\boxed{ \rm \:2cosx \: cosy = cos(x + y) + cos(x - y)}$

$\boxed{ \rm \:2sinx \: siny = cos(x - y) - cos(x + y)}$

$\boxed{ \rm \:cosx \: cos(60 \degree \: - x) \: cos(60 \degree \: + x) = \frac{1}{4} \: cos3x}$

$\boxed{ \rm \:sinx \: sin(60 \degree \: - x) \: sin(60 \degree \: + x) = \frac{1}{4} \: sin3x}$

$\boxed{ \rm \:sinx + siny = 2sin\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg]}$

$\boxed{ \rm \:sinx - siny = 2sin\bigg[\dfrac{x - y}{2} \bigg]cos\bigg[\dfrac{x + y}{2} \bigg]}$

$\boxed{ \rm \:cosx - cosy = - 2sin\bigg[\dfrac{x - y}{2} \bigg]sin\bigg[\dfrac{x + y}{2} \bigg]}$

$\boxed{ \rm \:cosx + cosy = 2cos\bigg[\dfrac{x- y}{2} \bigg]cos\bigg[\dfrac{x + y}{2} \bigg]}$