Question

prove that cos 2 theta* cos theta/2 - cos 3 theta *cos 9theta/2 = sin 5 theta sin 5theta/2

Answer 1

Step-by-step explanation:

Refer to the image below.

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Answer 2

Answer:

To Prove: $cos\,2\theta\times cos\,\frac{\theta}{2}-cos\,3\theta\times cos\,\frac{9\theta}{2}=sin\,5\theta\times sin\,\frac{5\theta}{2}$

We use the following result to prove  given statement:

$cos\,a \times cos\,b=\frac{1}{2}(cos\,(a+b)-cos\,(a-b))$

$cos\,a-cos\,b=-2sin\,(\frac{a+b}{2})\:sin\,(\frac{a-b}{2})$

Consider,

LHS

$=cos\,2\theta\times cos\,\frac{\theta}{2}-cos\,3\theta\times cos\,\frac{9\theta}{2}$

$=\frac{1}{2}(cos\,(2\theta+\frac{\theta}{2})-cos\,(2\theta-\frac{\theta}{2}))-\frac{1}{2}(cos\,(3\theta+\frac{9\theta}{2})-cos\,(3\theta-\frac{9\theta}{2}))$

$=\frac{1}{2}(cos\,\frac{5\theta}{2}-cos\,\frac{3\theta}{2}-(cos\,\frac{15\theta}{2}-cos\,\frac{-3\theta}{2}))$

$=\frac{1}{2}(cos\,\frac{5\theta}{2}-cos\,\frac{3\theta}{2}-cos\,\frac{15\theta}{2}+cos\,\frac{3\theta}{2})$

$=\frac{1}{2}(cos\,\frac{5\theta}{2}-cos\,\frac{15\theta}{2})$

$=\frac{1}{2}(-2sin\,\frac{20\theta}{4}\:sin\,\frac{5\theta}{2})$

$=sin\,5\theta\:sin\,\frac{5\theta}{2}$

= RHS

Hence Proved.