Expand \(\sin^5\theta\cos^3\theta\) in a series of sines of multiples of \(\theta\).

Let \(x=\cos\theta+i\sin\theta\): \(x-\frac1x=2i\sin\theta,\ x+\frac1x=2\cos\theta,\ x^n-\frac1{x^n}=2i\sin n\theta\).

\((2i\sin\theta)^5(2\cos\theta)^3=\left(x-\frac1x\right)^5\left(x+\frac1x\right)^3=\left(x-\frac1x\right)^2\left(x^2-\frac1{x^2}\right)^3\)

\(=\left(x^2-2+\frac1{x^2}\right)\left(x^6-3x^2+\frac3{x^2}-\frac1{x^6}\right)\)

\(=\left(x^8-\frac1{x^8}\right)-2\left(x^6-\frac1{x^6}\right)-2\left(x^4-\frac1{x^4}\right)+6\left(x^2-\frac1{x^2}\right)\)

LHS \(=32i\cdot8\sin^5\theta\cos^3\theta=256i\sin^5\theta\cos^3\theta\); RHS \(=2i\left[\sin8\theta-2\sin6\theta-2\sin4\theta+6\sin2\theta\right]\).

\(\sin^5\theta\cos^3\theta=\frac1{128}\left[\sin8\theta-2\sin6\theta-2\sin4\theta+6\sin2\theta\right]\) (verified numerically, e.g. \(\theta=0.7\) gives 0.04965 on both sides).