Expanding the sine and noting that i2=−1, we get: [1.2] Dividing by i, and grouping the sines: [1.3]
Bringing
the cosine to the end, in anticipation of an expansion, and taking cos
x out of the power to make it n-2k-2 (and multiplying by cos x): [1.4]
If n is even, let n=2p, so substituting in Equation 1.4 and making the cosine a square, with a view to using cos2x=1-sin2x
[1.5] If
n is odd, let n=2p-1, so the power of the cosine becomes 2p-2k-3, so to
make it a factor of 2, we multiply by the cos x, and drop it from the
equation: [1.6]
To accommodate both cases, we can write the final cosine as cosq x, where q is 0 when n is odd, and 1 when n is even: [1.7]
Writing (cos2 x)p-k-1 as 1−sin2 x)p-k-1 and writing the expansion: [1.8]
For a given k of the first sum, the k's will remain unchanged throughout, so we can group the powers of (-1) and of the sines: [1.9] Where p is the ceiling (n/2), or , and q=0, when n is odd, and q=1 when n is even. ■
A formula for the term with the power r of sine (that is, of sinrx) of sin nx is: [1.10] Where u=floor(r/2) and p=ceiling(n/2).
[2.1]
[1.1]
[1.2]
[1.3]
[1.4]
[1.5]
[1.6]
[1.7]
[1.8]
[1.9]
, and q=0, when n is odd, and q=1 when n is even. 
[1.10]