Question

Prove that $\sin (60-\theta ).\sin \theta .\sin (60+\theta )={\displaystyle \frac{\sin 3\theta }{4}}$.

Answer 1

Hint: Assume $\sin (60-\theta ).\sin \theta .\sin (60+\theta )=X$ and then use the trigonometric formula, $\sin (A+B)=\sin A\cos B+\cos A\sin B$ and $\sin (A-B)=\sin A\cos B-\cos A\sin B$ to solve the question.

Complete step-by-step answer:
We have been given to prove - $\sin (60-\theta ).\sin \theta .\sin (60+\theta )={\displaystyle \frac{\sin 3\theta }{4}}$.
Assuming $LHS=\sin (60-\theta ).\sin \theta .\sin (60+\theta )=X$
Therefore, $\sin (60-\theta ).\sin \theta .\sin (60+\theta )=X$.
Now using the trigonometric formula, $\sin (A+B)=\sin A\cos B+\cos A\sin B$ and $\sin (A-B)=\sin A\cos B-\cos A\sin B$, we can write-
$$\begin{gathered} \sin (60-\theta ).\sin \theta .\sin (60+\theta )=X \\ \Rightarrow (\sin {60}^{\circ }\cos \theta +\cos {60}^{\circ }\sin \theta ).\sin \theta .(\sin {60}^{\circ }\cos \theta -\cos {60}^{\circ }\sin \theta )=X \\ \Rightarrow ({\sin }^2{60}^{\circ }{\cos }^2\theta -{\cos }^2{60}^{\circ }{\sin }^2\theta )\sin \theta =X \\ \Rightarrow \left({\displaystyle \frac{3}{4}}{\cos }^2\theta -{\displaystyle \frac{{\sin }^2\theta }{4}}\right)\sin \theta =X\{\because \sin {60}^{\circ }={\displaystyle \frac{\sqrt{3}}{2}},\cos {60}^{\circ }={\displaystyle \frac{1}{2}}\} \\ \Rightarrow \left({\displaystyle \frac{3}{4}}(1-{\sin }^2\theta )-{\displaystyle \frac{{\sin }^2\theta }{4}}\right)\sin \theta =X \\ \Rightarrow \left({\displaystyle \frac{3}{4}}-{\sin }^2\theta \right)\sin \theta =X \\ \Rightarrow (3\sin \theta -4{\sin }^3\theta ){\displaystyle \frac{1}{4}}=X \\ \{\because 3\sin \theta -4{\sin }^3\theta =\sin 3\theta \} \\ \Rightarrow {\displaystyle \frac{\sin 3\theta }{4}}=X=RHS \end{gathered}$$
Therefore, LHS = RHS {Hence Proved}.

Note: Whenever such types of questions appear, first expand the term $\sin (60-\theta ),\sin (60+\theta )$ by using the trigonometric formula –$\sin (A-B)=\sin A\cos B-\cos A\sin B$ and the trigonometric formula $\sin (A+B)=\sin A\cos B+\cos A\sin B$, and then simplify the expression to prove it equal to the RHS.