Answer: Using the standard identity \(1+\cos\theta=2\cos^2\frac{\theta}{2}\) and \(\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\),

\(1+z=1+\cos\theta+i\sin\theta\)

\(=2\cos^2\frac{\theta}{2}+2i\sin\frac{\theta}{2}\cos\frac{\theta}{2}\)

\(=2\cos\frac{\theta}{2}\left(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}\right).\)

Also, by the binomial theorem,

\((1+z)^n=1+\binom{n}{1}z+\binom{n}{2}z^2+\cdots+\binom{n}{n}z^n.\)

Now \(z^k=(\cos k\theta+i\sin k\theta)\), so the imaginary part of \((1+z)^n\) is

\(\binom{n}{1}\sin\theta+\binom{n}{2}\sin2\theta+\cdots+\binom{n}{n}\sin n\theta.\)

On the other hand, from the identity for \(1+z\),

\((1+z)^n=\left(2\cos\frac{\theta}{2}\right)^n\left(\cos\frac{n\theta}{2}+i\sin\frac{n\theta}{2}\right).\)

So its imaginary part is

\(2^n\cos^n\frac{\theta}{2}\sin\frac{n\theta}{2}.\)

Hence

\(\binom{n}{1}\sin\theta+\binom{n}{2}\sin2\theta+\cdots+\binom{n}{n}\sin n\theta=2^n\cos^n\frac{\theta}{2}\sin\frac{n\theta}{2}.\)

Let \(z=\cos\theta+i\sin\theta\).

First, write

\(1+z=1+\cos\theta+i\sin\theta.\)

Use the half-angle identities

\(1+\cos\theta=2\cos^2\frac{\theta}{2}\), \quad \(\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}.\)

Then

\(1+z=2\cos^2\frac{\theta}{2}+2i\sin\frac{\theta}{2}\cos\frac{\theta}{2}\)

\(=2\cos\frac{\theta}{2}\left(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}\right).\)

This proves the first result.

Now consider \((1+z)^n\), where \(n\) is a positive integer. By the binomial theorem,

\((1+z)^n=1+\binom{n}{1}z+\binom{n}{2}z^2+\cdots+\binom{n}{n}z^n.\)

Since \(z^k=(\cos k\theta+i\sin k\theta)\), the imaginary part of \((1+z)^n\) is

\(\binom{n}{1}\sin\theta+\binom{n}{2}\sin2\theta+\cdots+\binom{n}{n}\sin n\theta.\)

On the other hand, from the factorisation found above,

\((1+z)^n=\left(2\cos\frac{\theta}{2}\right)^n\left(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}\right)^n.\)

Using de Moivre's theorem,

\(\left(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}\right)^n=\cos\frac{n\theta}{2}+i\sin\frac{n\theta}{2}.\)

So

\((1+z)^n=2^n\cos^n\frac{\theta}{2}\left(\cos\frac{n\theta}{2}+i\sin\frac{n\theta}{2}\right).\)

Therefore the imaginary part is

\(2^n\cos^n\frac{\theta}{2}\sin\frac{n\theta}{2}.\)

Equating imaginary parts gives

\(\binom{n}{1}\sin\theta+\binom{n}{2}\sin2\theta+\cdots+\binom{n}{n}\sin n\theta=2^n\cos^n\frac{\theta}{2}\sin\frac{n\theta}{2}.\)