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
 A: 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 $\sin(x)$
$${{\sin (7x)} \over {\sin (x)}} = 7 - 56\sin {\left( x \right)^2} + 112\sin {\left( x \right)^4} - 64\sin {\left( x \right)^6}$$
A: Take a look at Chebyshev Polynomial of the Second Kind.
Since $U_6 (\cos \theta) = \frac{\sin 7 \theta}{\sin \theta}$, we can find $U_6(x)$ be the recurrence given in the wikipedia link, i.e. $$U_0(x)=1$$ $$U_1(x)=2x$$ $$U_{n+1}(x)=2xU_n(x)-U_{n-1}(x)$$
We can then use $\cos^2 \theta = 1-\sin^2 \theta$ to convert it into a polynomial of $\sin$.
If you want to use DeMoivre's Theorem, use $$(\cos \theta + i \sin \theta)^7 = (\cos 7 \theta + i \sin 7 \theta)$$
Expand L.H.S to find the desired value.
A: $(\cos x+i\sin x)^7=\cos 7x+i\sin7x$.
We have
$$(r+t)^7=r^7+7r^6t+21r^5t^2+35r^4t^3+35r^3t^4+21r^2t^5+7rt^6+t^7$$
 hence, making $\sin x=t$ and $\cos x=r$ and having in account the powers of $i$, we get
 $$\sin 7x=7r^6t-35r^4t^3+21r^2t^5-t^7$$ so, because of $\cos^2x=1-\sin^2x$,
$$7t-21t^3+21t^5-7t^7-35t^3+70t^5-35t^7+21t^5-21t^7-t^7=-64t^7+112t^5-56t^3+7t$$
Now dividing by $t$, we finish.
A: $sin(7\theta)=sin(\theta+6\theta)=sin(\theta).cos(6\theta)+sin(6\theta)cos(\theta)$ 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.
