How to prove $\sin3θ=3\sinθ-4\sin^3θ$

Question

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!

Answer 1

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}

Answer 2

$$\begin{eqnarray} 3\sin\theta - 4\sin^3\theta &= 3\frac{e^{i\theta}-e^{-i\theta}}{2i}-4\frac{(e^{i\theta}-e^{-i\theta})^3}{8i^3}\\ &=\frac{e^{3i\theta}-(3-3)e^{i\theta}-(3-3)e^{-i\theta}-e^{-3i\theta}}{2i}=\\ &= \frac{e^{3i\theta}-e^{-3i\theta}}{2i} = \sin 3\theta \end{eqnarray}$$ where I used $(x+y)^3=x^3+3x^2y+3xy^2+y^3$, $i^2=-1$, $e^{a+b}=e^ae^b$, and $\sin(x)=(e^{ix}-e^{-ix})/2$.

Answer 3

\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*}

Answer 4

$$\sin(t+t+t)=\sin(t)\cos(t+t)+\cos(t)\sin(t+t)\\ =\sin(t)(\cos(t)\cos(t)-\sin(t)\sin(t))+\cos(t)(\sin(t)\cos(t)+\cos(t)\sin(t))\\ =\sin(t)(\cos^2(t)-\sin^2(t))+\cos(t)2\sin(t)\cos(t)\\ =\sin(t)(1-2\sin^2(t)+2\sin(t)(1-\sin^2(t))\\ =3\sin(t)-4\sin^3(t).$$

Answer 5

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.