show that `cos2thetacos(theta/2)-cos3thetacos(9theta/2)=sin5thetasin(5theta/2)`

Explanation

`LHS`



`=cos 2 theta cos frac{theta}{2}-cos 3 theta cos frac{9 theta}{2}
frac{1}{2}[2 cos 2 theta cos frac{theta}{2}-2 cos 3 theta cos frac{9 theta}{2}]`



(multiply divide equation by 2 )


[now, we know that `2 cos A cos B=cos (A+B)+cos (A-B)` ]



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