Question

Show that${\sin }^3\theta +{\sin }^3\left(\theta +120°\right)+{\sin }^3\left(\theta +240°\right)=-\frac{3}{4}\sin \left(3\theta \right)$

Short answer

To show that ${\sin }^3\theta +{\sin }^3\left(\theta +120°\right)+{\sin }^3\left(\theta +240°\right)=-\frac{3}{4}\sin \left(3\theta \right)$, take the LHS and simplify using the identity formulas.

Step-by-step solution

Step 1. Given information.

Consider the given question,

${\sin }^3\theta +{\sin }^3\left(\theta +120°\right)+{\sin }^3\left(\theta +240°\right)=-\frac{3}{4}\sin \left(3\theta \right)$

Take the LHS,

localid="1646423145211" $$\begin{gathered} ={\sin }^3\theta +{\sin }^3\left(\theta +120°\right)+{\sin }^3\left(\theta +240°\right) \\ ={\sin }^3\theta +{\left(\sin \theta \cos 120°+\cos \theta \sin 120°\right)}^3+{\left(\sin \theta \cos 240°+\cos \theta \sin 240°\right)}^3 \\ ={\sin }^3\theta +{\left(-\frac{1}{2}\sin \theta +\frac{\sqrt{3}}{2}\cos \theta \right)}^3+{\left(-\frac{1}{2}\sin \theta -\frac{\sqrt{3}}{2}\cos \theta \right)}^3 \\ ={\sin }^3\theta +\left(-\sin \theta \right)\left(\frac{1}{4}{\sin }^2\theta +\frac{3}{4}{\cos }^2\theta -\frac{1}{4}{\sin }^2\theta +\frac{3}{4}{\cos }^2\theta \right) \end{gathered}$$

Step 2. Use the identity formulas and simplify the expression.

Continuing the above expression,

$$\begin{gathered} =\frac{3}{4}{\sin }^3\theta -\frac{9}{4}{\cos }^2\theta \sin \theta \\ ={\sin }^3\theta +\left(-\frac{{\sin }^3\theta }{4}-\frac{9}{4}{\cos }^2\theta \sin \theta \right) \end{gathered}$$

Use the identity ${\cos }^2\theta =1-{\sin }^2\theta$,

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