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!
 A: 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}
A: $$\begin{eqnarray}
3\sin\theta - 4\sin^3\theta &= 3\frac{e^{i\theta}-e^{-i\theta}}{2i}-4\frac{(e^{i\theta}-e^{-i\theta})^3}{8i^3}\\
 &=\frac{e^{3i\theta}-(3-3)e^{i\theta}-(3-3)e^{-i\theta}-e^{-3i\theta}}{2i}=\\
&= \frac{e^{3i\theta}-e^{-3i\theta}}{2i} = \sin 3\theta
\end{eqnarray}$$
where I used $(x+y)^3=x^3+3x^2y+3xy^2+y^3$, $i^2=-1$, $e^{a+b}=e^ae^b$, and $\sin(x)=(e^{ix}-e^{-ix})/2$.
A: \begin{align*}
\sin 3\theta&=\sin(\theta+2\theta)\\
&=\sin \theta\cos 2\theta+\cos\theta\sin2\theta\\
&=\sin\theta(\cos^2\theta-\sin^2\theta)+2\cos^2\theta\sin\theta\\
&=3\cos^2\theta\sin\theta-\sin^3\theta\\
&=3(1-\sin^2\theta)\sin\theta-\sin^3\theta\\
&=3\sin\theta-4\sin^3\theta
\end{align*}
A: $$\sin(t+t+t)=\sin(t)\cos(t+t)+\cos(t)\sin(t+t)\\
=\sin(t)(\cos(t)\cos(t)-\sin(t)\sin(t))+\cos(t)(\sin(t)\cos(t)+\cos(t)\sin(t))\\
=\sin(t)(\cos^2(t)-\sin^2(t))+\cos(t)2\sin(t)\cos(t)\\
=\sin(t)(1-2\sin^2(t)+2\sin(t)(1-\sin^2(t))\\
=3\sin(t)-4\sin^3(t).$$
A: Basic approach. First write $\sin 3\theta$ as $\sin (\theta+2\theta)$, and use the sine of sum formula.  Then write $\cos 2\theta$ as $1-2\sin^2\theta$ and $\sin 2\theta$ as $2\sin\theta\cos\theta$.  From there, use $\cos^2\theta = 1-\sin^2\theta$, and simple algebra gets you the rest of the way.
