Question

How do you express $\sin \left(3\theta \right)$ in terms of trigonometric functions of $\theta$?

Answer 1

Finding the required formula:

As we know,

$$\begin{gathered} \sin (x+y)=\sin x\cos y+\cos x\sin y \\ \cos (x+y)=\cos x\cos y–\sin x\sin y \end{gathered}$$

We can write that

$\begin{array}{rcl}\sin \left(3\theta \right)& =& \sin (2\theta +\theta )\\ & =& \sin \left(2\theta \right)\cos \left(\theta \right)+\cos \left(2\theta \right)\sin \left(\theta \right)\\ & =& \sin (\theta +\theta )\cos \left(\theta \right)+\cos (\theta +\theta )\sin \left(\theta \right)\\ & =& \left[\sin \right(\theta )\cos (\theta )+\cos (\theta )\sin (\theta \left)\right]\cos \left(\theta \right)+\left[\cos \right(\theta )\cos \left(\theta \right)+\sin \left(\theta \right)\sin \left(\theta \right)]\sin \left(\theta \right)\\ & =& 2\sin \left(\theta \right){\cos }^2\left(\theta \right)+\sin \left(\theta \right){\cos }^2\left(\theta \right)-{\sin }^3\left(\theta \right)\\ & =& 3\sin \left(\theta \right){\cos }^2\left(\theta \right)-{\sin }^3\left(\theta \right)\end{array}$

$$\begin{gathered} =3\sin \left(\theta \right)(1-{\sin }^2(\theta \left)\right)-{\sin }^3\left(\theta \right) \\ =3\sin \left(\theta \right)-3{\sin }^3\left(\theta \right)-{\sin }^3\left(\theta \right) \\ =3\sin \left(\theta \right)-4{\sin }^3\left(\theta \right) \end{gathered}$$ $\left[{\sin }^2\theta +{\cos }^2\theta =1\right]$

Hence, the value of $\sin \left(3\theta \right)$ is $3\sin \left(\theta \right)-4{\sin }^3\left(\theta \right)$.