If \(\cos\alpha+\cos\beta+\cos\gamma=0,\ \sin\alpha+\sin\beta+\sin\gamma=0\), prove that (i) \(\cos3\alpha+\cos3\beta+\cos3\gamma=3\cos(\alpha+\beta+\gamma)\) (ii) \(\sin3\alpha+\sin3\beta+\sin3\gamma=3\sin(\alpha+\beta+\gamma)\)

Let \(a=\cos\alpha+i\sin\alpha,\ b=\cos\beta+i\sin\beta,\ c=\cos\gamma+i\sin\gamma\).

Then \(a+b+c=(\cos\alpha+\cos\beta+\cos\gamma)+i(\sin\alpha+\sin\beta+\sin\gamma)=0\).

Identity: if \(a+b+c=0\) then \(a^3+b^3+c^3=3abc\) (since \(a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\)).

By De Moivre: \(a^3=\cos3\alpha+i\sin3\alpha\), etc., and \(abc=\cos(\alpha+\beta+\gamma)+i\sin(\alpha+\beta+\gamma)\).

\(\therefore(\cos3\alpha+\cos3\beta+\cos3\gamma)+i(\sin3\alpha+\sin3\beta+\sin3\gamma)=3\cos(\alpha+\beta+\gamma)+3i\sin(\alpha+\beta+\gamma)\)

Equating real and imaginary parts: (i) \(\cos3\alpha+\cos3\beta+\cos3\gamma=3\cos(\alpha+\beta+\gamma)\); (ii) \(\sin3\alpha+\sin3\beta+\sin3\gamma=3\sin(\alpha+\beta+\gamma)\) (proved).