Answer: \(\binom{n}{1}\cos\theta+\binom{n}{2}\cos2\theta+\cdots+\binom{n}{n}\cos n\theta=2^n\cos^n\left(\frac\theta2\right)\cos\left(\frac{n\theta}{2}\right)-1\).
\(\binom{n}{1}\sin\theta+\binom{n}{2}\sin2\theta+\cdots+\binom{n}{n}\sin n\theta=2^n\cos^n\left(\frac\theta2\right)\sin\left(\frac{n\theta}{2}\right)\).
By the binomial theorem,
\((1+z)^n=1+\binom{n}{1}z+\binom{n}{2}z^2+\cdots+\binom{n}{n}z^n\).
Since \(z=\cos\theta+i\sin\theta\), de Moivre's theorem gives \(z^k=\cos k\theta+i\sin k\theta\). Therefore the real part is
\(\operatorname{Re}\{(1+z)^n\}=1+\binom{n}{1}\cos\theta+\binom{n}{2}\cos2\theta+\cdots+\binom{n}{n}\cos n\theta\).
Now
\(1+z=1+\cos\theta+i\sin\theta\).
Using \(1+\cos\theta=2\cos^2\left(\frac\theta2\right)\) and \(\sin\theta=2\sin\left(\frac\theta2\right)\cos\left(\frac\theta2\right)\),
\(1+z=2\cos\left(\frac\theta2\right)\left(\cos\left(\frac\theta2\right)+i\sin\left(\frac\theta2\right)\right)\).
Hence
\((1+z)^n=2^n\cos^n\left(\frac\theta2\right)\left(\cos\left(\frac\theta2\right)+i\sin\left(\frac\theta2\right)\right)^n\).
By de Moivre's theorem,
\((1+z)^n=2^n\cos^n\left(\frac\theta2\right)\left(\cos\left(\frac{n\theta}{2}\right)+i\sin\left(\frac{n\theta}{2}\right)\right)\).
Equating real parts gives
\(1+\binom{n}{1}\cos\theta+\binom{n}{2}\cos2\theta+\cdots+\binom{n}{n}\cos n\theta=2^n\cos^n\left(\frac\theta2\right)\cos\left(\frac{n\theta}{2}\right)\).
Therefore
\(\binom{n}{1}\cos\theta+\binom{n}{2}\cos2\theta+\cdots+\binom{n}{n}\cos n\theta=2^n\cos^n\left(\frac\theta2\right)\cos\left(\frac{n\theta}{2}\right)-1\).
Equating imaginary parts instead gives
\(\binom{n}{1}\sin\theta+\binom{n}{2}\sin2\theta+\cdots+\binom{n}{n}\sin n\theta=2^n\cos^n\left(\frac\theta2\right)\sin\left(\frac{n\theta}{2}\right)\).
