Prove that \(\cos^4\theta\sin^3\theta=-\frac{1}{64}\left(\sin7\theta+\sin5\theta-3\sin3\theta-3\sin\theta\right)\).

Answer: Proved, using \(x=\cos\theta+i\sin\theta\) so that \(2\cos\theta=x+\frac1x\) and \(2i\sin\theta=x-\frac1x\).

Let \(x=\cos\theta+i\sin\theta\). Then \(\frac1x=\cos\theta-i\sin\theta\), and by De Moivre's theorem

\[x^n+\frac1{x^n}=2\cos n\theta,\qquad x^n-\frac1{x^n}=2i\sin n\theta.\]

Left side: \((2\cos\theta)^4(2i\sin\theta)^3=2^7\,i^3\cos^4\theta\sin^3\theta=-128\,i\cos^4\theta\sin^3\theta\).

Right side in powers of \(x\):

\[\left(x+\tfrac1x\right)^4\left(x-\tfrac1x\right)^3=\left(x^2-\tfrac1{x^2}\right)^3\left(x+\tfrac1x\right)\]

\[=\left(x^6-3x^2+\tfrac{3}{x^2}-\tfrac1{x^6}\right)\left(x+\tfrac1x\right)=x^7+x^5-3x^3-3x+\tfrac3x+\tfrac3{x^3}-\tfrac1{x^5}-\tfrac1{x^7}\]

\[=\left(x^7-\tfrac1{x^7}\right)+\left(x^5-\tfrac1{x^5}\right)-3\left(x^3-\tfrac1{x^3}\right)-3\left(x-\tfrac1x\right)\]

\[=2i\left(\sin7\theta+\sin5\theta-3\sin3\theta-3\sin\theta\right).\]

Equate: \(-128\,i\cos^4\theta\sin^3\theta=2i(\sin7\theta+\sin5\theta-3\sin3\theta-3\sin\theta)\)

\[\therefore\ \cos^4\theta\sin^3\theta=-\frac{1}{64}\left(\sin7\theta+\sin5\theta-3\sin3\theta-3\sin\theta\right).\]

Quick check: at \(\theta=\pi/2\): LHS \(=0\); RHS \(=-\frac1{64}(-1+1+3-3)=0\). ✓