Question

prove cos theta cos theta/2 -cos 3 theta cos 9theta/2 = sin 4 theta * sin 7 theta/2

Answer 1

Question :- Prove that

$\rm \: cos\theta \: cos\dfrac{\theta }{2} - cos3\theta \: cos\dfrac{9\theta }{2} = sin4\theta \: sin\dfrac{7\theta }{2}$

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

Consider LHS

$\rm \: cos\theta \: cos\dfrac{\theta }{2} - cos3\theta \: cos\dfrac{9\theta }{2}$

can be rewritten as

$\rm \: = \dfrac{1}{2} \bigg(2\: cos\theta \: cos\dfrac{\theta }{2} -2 cos3\theta \: cos\dfrac{9\theta }{2}\bigg)$

We know,

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

So, using this result, we get

$\rm \:=\:\dfrac{1}{2}\bigg[cos\bigg(\theta + \dfrac{\theta }{2} \bigg) + cos\bigg(\theta - \dfrac{\theta }{2} \bigg) - cos\bigg(3\theta + \dfrac{9\theta }{2} \bigg) - cos\bigg(3\theta - \dfrac{9\theta }{2}\bigg) \bigg]$

$\rm \:=\: \dfrac{1}{2}\bigg[cos\dfrac{3\theta }{2} + cos\dfrac{\theta }{2} - cos\dfrac{15\theta }{2} - cos\bigg( - \dfrac{3\theta }{2} \bigg) \bigg]$

$\rm \:=\: \dfrac{1}{2}\bigg[cos\dfrac{3\theta }{2} + cos\dfrac{\theta }{2} - cos\dfrac{15\theta }{2} - cos \dfrac{3\theta }{2}\bigg]$

$\rm \:=\: \dfrac{1}{2}\bigg[ cos\dfrac{\theta }{2} - cos\dfrac{15\theta }{2} \bigg]$

We know,

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

So, using this result, we get

$\rm \: = \: \dfrac{1}{2} \times 2sin\bigg[\dfrac{\dfrac{\theta }{2} + \dfrac{15\theta }{2}}{2} \bigg]sin\bigg[\dfrac{\dfrac{15\theta }{2} - \dfrac{\theta }{2}}{2} \bigg]$

$\rm \: = \: sin\dfrac{16\theta }{4} \: sin\dfrac{14\theta }{4}$

$\rm \: = \: sin4\theta \: sin\dfrac{7\theta }{2}$

Hence,

$\rm\implies \:\boxed{\rm{  \:\rm \: cos\theta \: cos\dfrac{\theta }{2} - cos3\theta \: cos\dfrac{9\theta }{2} = sin4\theta \: sin\dfrac{7\theta }{2} \: }}$

$\rule{190pt}{2pt}$

$\begin{gathered} \colorbox{powderblue}{ \boxed{ \begin{array}{l} \underline{\underline{ \color{orang} \text{ \bf \: Additional \: information \: }}} \end{array}}}\end{gathered}$

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

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

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

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

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