Find $(\sin x)^7$, reduced in specific terms $(\sin x)^7 = a\sin 7x+b\sin 5x+c\sin 3x+d\sin x$. Find $d$. $x$ is an angle, and $a, b, c, d$ are all constants. 
I am not sure where to start! The only idea I have is to possibly break up $(\sin x)^7$ into smaller parts such as $(\sin x)^2 * (\sin x)^2 * (\sin x)^3$. 
Hints only, please. 
 A: Hint.  Using Euler's formula and related things,
$$\eqalign{
  (\sin x)^7
  &=\Bigl(\frac{e^{ix}-e^{-ix}}{2i}\Bigr)^7\cr
  &=\cdots\quad\langle\hbox{expand by binomial theorem}\rangle\cr
  &=\frac{1}{(2i)^7}(e^{7ix}-e^{-7ix}+\langle\hbox{more terms}\rangle)\cr
  &=-\frac{1}{64}\sin7x+\langle\hbox{more terms}\rangle\ .\cr}$$
See if you can do the rest.
A: We do a much simpler problem, which can be a step in the solution of your problem. The key is the product to sum formula
$$\sin a\cos b=\frac{\sin(a+b)+\sin(a-b)}{2}.\tag{1}$$
We also use the double-angle formula
$$\sin^2 x=\frac{1-\cos(2x)}{2}.\tag{2}$$ 
Now we do $(\sin x)^3$. This is $\frac{\sin x}{2}-\frac{\sin x\cos 2x}{2}$. Now formula (1) gives $\frac{\sin x}{2}-\frac{1}{4}\left(\sin(3x)+\sin(-x)\right)$, which 
simplifies to 
$$\frac{1}{4}\left(3\sin x-\sin(3x)\right).\tag{3}$$
For $(\sin x)^5$, multiply our expression by $\sin^2 x$, that is, the right-hand side of (2), and use sum to product again.
Then comes $(\sin x)^7$.
Remark: Complex numbers are the right thing to use. But the sketch above shows it can be done using standard trigonometric identities. However, these trigonometric identities hold "because" of the more natural fact that $\exp(w)\exp(z)=\exp(z+w)$.  
