Some trigonometry formula (IV)

Last time we proved $ \sin 3x = \frac{1}{4}(3\sin x - \sin 3x) $ and we promised to go for the $ \sin^n x $, so here we are, using recurrences.

We start with:

$ \sin^{n - 2} x = \sum a_j \sin jx $

Next we multiply $ \sin^2 x $ on both side and get

$ \begin{eqnarray*} \sin^{n - 2} x &=& \sum a_j \sin jx \\ \sin^n x &=& \sum a_j \sin^2 x \sin jx \\ &=& \frac{1}{2}\sum a_j (\sin x (\cos (x - jx)) - \cos (x + jx))) \\ &=& \frac{1}{2}\sum a_j (\sin x \cos (x - jx)) - \sin x \cos (x + jx)) \\ &=& \frac{1}{4}\sum a_j (\sin (x + x - jx) + \sin (x - (x - jx)) - \sin (x + x + jx) - \sin (x - (x + jx))) \\ &=& \frac{1}{4}\sum a_j (\sin (2 - j)x + \sin (jx) - \sin (j + 2)x - \sin (-jx)) \\ &=& \frac{1}{4}\sum a_j (-\sin (j - 2)x + \sin (jx) - \sin (j + 2)x + \sin (jx)) \\ &=& \frac{1}{4}\sum a_j (-\sin (j - 2)x + 2\sin (jx) - \sin (j + 2)x) \\ \end{eqnarray*} $

So that's the theory. Imagine this, take every term in the sine series, push a negative value to your neighbors, and double itself. Sum these little pieces together

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

We apply the principle once again, we get

1 -2 1
  -2 4 -2
     1 -2 1

$ \sin^5 x = \frac{1}{16}(\sin 5x - 4\sin(3x) + 6\sin x - 4 \sin(-x) + \sin(-3x)) $

Note the beautiful symmetry and binomial coefficients!