Question

$\frac{\sin 3\theta +2\sin 5\theta +\sin 7\theta }{\sin \theta +2\sin 3\theta +\sin 5\theta }=\cos 2\theta +\sin 2\theta .\cot 3\theta$.

Answer 1

$\frac{sin3\theta +2sin\theta +sin7\theta }{sin\theta +2sin3\theta +sin5\theta }$

$\therefore \frac{sin3\theta +sin7\theta +2sin5\theta }{sin\theta +sin5\theta +2sin3\theta }$

$=\frac{2sin5\theta cos2\theta +2sin5\theta }{2sin3\theta cos2\theta +2sin3\theta }.....(sinA+sinB=2sin\frac{(A+B)}{2}cos\frac{(A-B)}{2})$

$=\frac{2sin5\theta (cos2\theta +1)}{2sin3\theta (cos2\theta +1)}$

$=\frac{sin5\theta }{sin3\theta }\to LHS$

Now, RHS , $cos2\theta +sin2\theta cot3\theta$

$=cos2\theta +sin2\theta \times \frac{cos3\theta }{sin3\theta }$

$\frac{sin3\theta cos2\theta +sin2\theta cos3\theta }{sin3\theta }$

$=\frac{sin5\theta }{sin3\theta }......\left(sin\right(A+B)=sinAcosB+cosAsinB)$

$=LHS$

Hence LHS=RHS

Hence proved.