Answer
For $cos3\theta$, equate the real part from both sides
$cos3\theta=cos^{3}\theta-3sin^{2}\theta cos\theta$
For $sin3\theta$, equate the real part from both sides
$sin3\theta=3sin\theta cos^{2}\theta-sin^{3}\theta$
Work Step by Step
Consider $z=cos\theta+isin\theta$
Cubing this we have,
$(cos\theta+isin\theta)^{3}=cos^{3}\theta-3sin^{2}\theta cos\theta+i(3sin\theta cos^{2}\theta-sin^{3}\theta)$
Apply De-Moivre's theorem on the left side:
$z^{n}=r^{n}(cosn\theta+isinn\theta)$
we have
$1^{3}(cos3\theta+isin3\theta)=cos^{3}\theta-3sin^{2}\theta cos\theta+i(3sin\theta cos^{2}\theta-sin^{3}\theta)$
$(cos3\theta+isin3\theta)=cos^{3}\theta-3sin^{2}\theta cos\theta+i(3sin\theta cos^{2}\theta-sin^{3}\theta)$
For $cos3\theta$, equate the real part from both sides
$cos3\theta=cos^{3}\theta-3sin^{2}\theta cos\theta$
For $sin3\theta$, equate the real part from both sides
$sin3\theta=3sin\theta cos^{2}\theta-sin^{3}\theta$
