# How do you prove sin 3 theta = 3 sin theta - 4 sin^3 theta?

Jun 20, 2016

see explanation

#### Explanation:

Express the left hand side as

$\sin 3 \theta = \sin \left(\theta + 2 \theta\right)$

now expand the right side of this equation using $\textcolor{b l u e}{\text{Addition formula}}$

color(red)(|bar(ul(color(white)(a/a)color(black)(sin(A±B)=sinAcosB±cosAsinB)color(white)(a/a)|)))

$\Rightarrow \sin \left(\theta + 2 \theta\right) = \sin \theta \cos 2 \theta + \cos \theta \sin 2 \theta \ldots \ldots . \left(A\right)$

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\cos 2 \theta = {\cos}^{2} \theta - {\sin}^{2} \theta = 2 {\cos}^{2} \theta - 1 = 1 - 2 {\sin}^{2} \theta} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The right hand side is expressed only in terms of $\sin \theta ' s$

so we use $\cos 2 \theta = 1 - 2 {\sin}^{2} \theta \ldots \ldots . . \left(1\right)$

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\sin 2 \theta = 2 \sin \theta \cos \theta} \textcolor{w h i t e}{\frac{a}{a}} |}}} \ldots \ldots . . \left(2\right)$

Replace $\cos 2 \theta \text{ and } \sin 2 \theta$ by the expansions (1) and (2)
into (A)

sin(theta+2theta)=sinthetacolor(red)((1-2sin^2theta))+costhetacolor(red)((2sinthetacostheta)

and expanding brackets gives.

$\sin \left(\theta + 2 \theta\right) = \sin \theta - 2 {\sin}^{3} \theta + 2 \sin \theta {\cos}^{2} \theta \ldots . \left(B\right)$

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\cos}^{2} \theta + {\sin}^{2} \theta = 1 \Rightarrow {\cos}^{2} \theta = 1 - {\sin}^{2} \theta} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Replace ${\cos}^{2} \theta = 1 - {\sin}^{2} \theta \text{ into (B)}$

$\Rightarrow \sin \left(\theta + 2 \theta\right) = \sin \theta - 2 {\sin}^{3} \theta + 2 \sin \theta \left(1 - {\sin}^{2} \theta\right)$

and expanding 2nd bracket gives.

$\sin \left(\theta + 2 \sin \theta\right) = \sin \theta - 2 {\sin}^{3} \theta + 2 \sin \theta - 2 {\sin}^{3} \theta$

Finally, collecting like terms.

$\sin 3 \theta = 3 \sin \theta - 4 {\sin}^{3} \theta = \text{R.H.S hence proven}$