Express $\frac{\sin 7\theta}{\sin \theta}$ in powers of $\sin \theta$ only By using DeMoivre's theorm express
$$\frac{\sin 7\theta}{\sin \theta}$$
in the powers of Sine only

answer given in the book is
$$7-56\sin ^2\theta+112\sin ^4 \theta-64\sin^6 \theta$$
can any one help to solve the question
Steps To Carry Out

1) First take a look at this link which is a guide for DeMoivre's formula.

2) Using step 1 show that

$$\sin (7x) = 64\sin \left( x \right)\cos {\left( x \right)^6} - 80\sin \left( x \right)\cos {\left( x \right)^4} + 24\sin \left( x \right)\cos {\left( x \right)^2} - \sin \left( x \right)$$

3) Replace $\cos^2(x)=1-\sin^2(x)$ and obtain

$$\sin (7x) = 7\sin \left( x \right) - 56\sin {\left( x \right)^3} + 112\sin {\left( x \right)^5} - 64\sin {\left( x \right)^7}$$

4) Divide by $\sin(x)$

$${{\sin (7x)} \over {\sin (x)}} = 7 - 56\sin {\left( x \right)^2} + 112\sin {\left( x \right)^4} - 64\sin {\left( x \right)^6}$$
$sin(7\theta)=sin(\theta+6\theta)=sin(\theta).cos(6\theta)+sin(6\theta)cos(\theta)$ then we have $cos(6\theta)=cos(3\theta+3\theta)=cos(3\theta).cos(\theta)-[sin(3\theta).sin(3\theta)]$ now we know $cos(3\theta)=4cos^3(\theta)-3cos(\theta),sin(\theta)=3sin(\theta)+4sin^3(\theta)$. Now you can just back substitute and get the job done. Hope its clear.