Some trigonometry formula (III)

Now we are starting to do something fun out of what we derived. The goal is to find out an expression for $\sin^n x$ for odd $n$ in terms of $\sin k x$ only.

Obviously, we have a base case for $n = 1$, that give $\sin x = \sin x$, not very interesting, I know.

How about $\sin^3 x$? We can start using the product to sum formula we derived.


$\begin{eqnarray*} & & \sin^3 x \\ &=& \sin x (\sin x (\sin x)) \\ &=& \sin x (\frac{1}{2}(\cos(x - x) - \cos(x + x))) \\ &=& \sin x (\frac{1}{2}(\cos(0x) - \cos(2x))) \\ &=& \frac{1}{2}(\sin x \cos(0x) - \sin x \cos(2x)) \\ &=& \frac{1}{2}(\frac{1}{2}(\sin(x + 0x) + \sin(x - 0x)) - \frac{1}{2}(\sin(x + 2x) + \sin(x - 2x))) \\ &=& \frac{1}{4}(\sin(x + 0x) + \sin(x - 0x) - \sin(x + 2x) - \sin(x - 2x)) \\ &=& \frac{1}{4}(\sin(x) + \sin(x) - \sin(3x) - \sin(-x)) \\ &=& \frac{1}{4}(\sin(x) + \sin(x) - \sin(3x) + \sin(x)) \\ &=& \frac{1}{4}(3\sin(x) - \sin(3x)) \\ \end{eqnarray*}$

You probably wondered, why do I keep all those $\cos 0x$ around. That is done to show that exactly the same trick can be applied to general $n$.

Let's do that in the next post!