Use DeMoivre's formula to derive the trig identities $$\begin{cases} \cos3\theta = (\cos\theta)^2-3\cos\theta(\sin\theta)^2\\ \sin3\theta=3(\cos\theta)^2\sin\theta-(\sin\theta)^3 \end{cases}$$ and note: $(a+b)^3 = a^3+3a^2b+3ab^2+b^3$
I'm kind of clueless on how to do this. I missed class when we learned how to apply DeMoivre's formula.
DeMoivre's formula looks like this: $$ \color{maroon}{\cos(n\theta)} + i\color{navy}{\sin(n\theta)} = (\cos(\theta) + i\sin(\theta))^n $$ Use the case where $n=3$ and multiply out the RHS. So $\color{maroon}{\cos(3\theta)} + i\color{navy}{\sin(3\theta)} = $ $$ \begin{align} &= (\cos(\theta) + i\sin(\theta))^3 \\ &=\big(\cos^2(\theta)-\sin^2(\theta)+2i\sin(\theta)\cos(\theta)\big)(\cos(\theta) + i\sin(\theta)) \\ &=\cos^3(\theta)-\sin^2(\theta)\cos(\theta)+2i\sin(\theta)\cos^2(\theta)+i\sin(\theta)\cos^2(\theta)-i\sin^3(\theta)-2\sin^2(\theta)\cos(\theta)\\ &= \dotsc \end{align} $$ And then we can group the terms with a coefficient of $i$ and those without a coefficient of $i$: $$ \begin{align} \dotsc &= \color{maroon}{\big(\cos^3(\theta)-3\sin^2(\theta)\cos(\theta)\big)} + i\color{navy}{\big(-\sin^3(\theta)+3\sin(\theta)\cos^2(\theta)\big)} \end{align} $$ Then looking back at the $\color{maroon}{\cos(3\theta)} + i\color{navy}{\sin(3\theta)}$ that we started with, we can see that the $\color{maroon}{\text{real term}}$ in this expression must be equal to the $\color{maroon}{\text{real term}}$ in our result, and the $\color{navy}{\text{imaginary term}}$ in this expression must be equal to the $\color{navy}{\text{imaginary term}}$ in our result. So, $$ \begin{cases}\color{maroon}{\cos(3\theta) = \cos^3(\theta)-3\sin^2(\theta)\cos(\theta)}\\ \color{navy}{\sin(3\theta) = -\sin^3(\theta)+3\sin(\theta)\cos^2(\theta)} \end{cases} $$