Question

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

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 - $\sin (60 - \theta ).\sin \theta .\sin (60 + \theta ) = \dfrac{{\sin 3\theta }}{4}$.
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-
$
  \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( {\dfrac{3}{4}{{\cos }^2}\theta - \dfrac{{{{\sin }^2}\theta }}{4}} \right)\sin \theta = X\{ \because \sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2},\cos {60^ \circ } = \dfrac{1}{2}\} \\
   \Rightarrow \left( {\dfrac{3}{4}(1 - {{\sin }^2}\theta ) - \dfrac{{{{\sin }^2}\theta }}{4}} \right)\sin \theta = X \\
   \Rightarrow \left( {\dfrac{3}{4} - {{\sin }^2}\theta } \right)\sin \theta = X \\
   \Rightarrow (3\sin \theta - 4{\sin ^3}\theta )\dfrac{1}{4} = X \\
  \{ \because 3\sin \theta - 4{\sin ^3}\theta = \sin 3\theta \} \\
   \Rightarrow \dfrac{{\sin 3\theta }}{4} = X = RHS \\
$
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.