<s>[INST] How to prove $\sin3θ=3\sinθ-4\sin^3θ$ I was solving an A Level past paper (November 2014 P32) when I stumbled upon this question. It first asks us to expand $\sin(2θ+θ)$ which is easy using the identity $\sin(A+B)=\sin A\cos B+\cos A\sin B$, but what I am not able to do is prove that this is equal to $3\sin θ-4\sin^3 θ$. I tried using the $\sin 2A=2 \sin A \cos A$ identity and the basic $\cos^2θ=1-\sin^2θ$ identity, but that isn't working. I'd really appreciate it if someone would guide me through this question. Thanks in advance!
 [/INST] We will use the following trigonometric formulas:
\begin{align}
\color{red}{\sin(x+y)\,}&\color{red}{=\sin x\cos y+\sin y\cos x}\\
\color{green}{\sin (2x)\,}&\color{green}{=2\sin x\cos x}\\
\color{blue}{\cos (2x)\,}&\color{blue}{=1-2\sin^2x}\\
\color{magenta}{\cos^2 x\,}&\color{magenta}{=1-\sin^2 x}
\end{align}
So
\begin{align}
\sin(2\theta+\theta)\,&=\color{red}{\sin(2\theta)\cos \theta+\sin \theta\cos(2\theta)}\\
&=\color{green}{2\sin\theta\cos^2\theta}+\sin \theta(\color{blue}{1-2\sin^2\theta})\\
&=2\sin\theta(\color{magenta}{1-\sin^2\theta})+\sin\theta-2\sin^3\theta\\
&=3\sin\theta-4\sin^3\theta
\end{align}
</s>