Question

Prove that $\sin 3x+\sin 2x-\sin x=4\sin x\cos \dfrac{x}{2}\cos \dfrac{3x}{2}$.

Answer 1

Hint: Use the trigonometric formulas of $\left( \sin A-\sin B \right)$, value of $\sin 2x$to solve the expression. Take LHS of the expression and prove that the simplified expression of LHS is equal to RHS.


Complete step-by-step solution -
Given the expression, $\sin 3x+\sin 2x-\sin x=4\sin x\cos \dfrac{x}{2}\cos \dfrac{3x}{2}$.
Let us first take the LHS.
$LHS =\sin 3x +\sin 2x-\sin x$.
We know the formula, $\sin A-\sin B=2\cos \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)$
Let us put, A = 2x and B = x.
$\therefore \sin 3x+\left( \sin 2x-\sin x \right)=\sin 3x+2\cos \left( \dfrac{2x+x}{2} \right)\sin \left( \dfrac{2x-x}{2} \right)$
                                                 $=\sin 3x+2\cos \left( \dfrac{3x}{2} \right)\sin \left( \dfrac{x}{2} \right)$
We know that, $\sin 2x=2\sin x.\cos x$.
Similarly, $\sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$.
Now let us replace x with 3x.
$\begin{align}
  & \therefore \sin 3x=2\sin \left( \dfrac{3x}{2} \right)\cos \left( \dfrac{3x}{2} \right) \\
 & \therefore LHS=2\sin \left( \dfrac{3x}{2} \right)\cos \left( \dfrac{3x}{2} \right)+2\cos \left( \dfrac{3x}{2} \right)\sin \left( \dfrac{x}{2} \right) \\
\end{align}$
              $=2\cos \left( \dfrac{3x}{2} \right)\left[ \sin \left( \dfrac{3x}{2} \right)+\sin \left( \dfrac{x}{2} \right) \right]$
We know that, $\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$.
Put, $A=\dfrac{3x}{2}$and $B=\dfrac{x}{2}$.
$\therefore LHS=2\cos \left( \dfrac{3x}{2} \right)\left[ 2\sin \left( \dfrac{\dfrac{3x}{2}+\dfrac{x}{2}}{2} \right)\cos \left( \dfrac{\dfrac{3x}{2}-\dfrac{x}{2}}{2} \right) \right]$
              $\begin{align}
  & =2\cos \left( \dfrac{3x}{2} \right)\left[ 2\sin \left( \dfrac{4x}{4} \right)\cos \left( \dfrac{2x}{4} \right) \right] \\
 & =2\cos \left( \dfrac{3x}{2} \right)\left[ 2\sin x\cos \left( \dfrac{x}{2} \right) \right] \\
 & =4\cos \left( \dfrac{3x}{2} \right)\sin x\cos \dfrac{x}{2} \\
 & =4\sin x\cos \left( \dfrac{x}{2} \right)\cos \left( \dfrac{3x}{2} \right) \\
\end{align}$
Hence, we proved that LHS = RHS.
Here from the question, $RHS=4\cos \left( \dfrac{x}{2} \right)\sin x\cos \left( \dfrac{3x}{2} \right)$.
Thus, we have proved.

Note: Remember the trigonometric formulae and identities to solve problems like these. Take $\left( \sin 2x-\sin x \right)$first because you will get a simplified expression and will be able to solve the expression more easily. We have used formulas like $\left( \sin A-\sin B \right)$, $\sin 2x,\left( \sin A+\sin B \right)$ to solve the expression. Therefore, remember the formulae.