Question

Prove the following:
${{\sin }^{2}}6x-{{\sin }^{2}}4x=\sin 2x\sin 10x$

Answer 1

Hint: For solving this question, we will simplify the term on the left-hand side and prove that it is equal to the term on the right-hand side. And we will use formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ and trigonometric formulas of $\sin C+\sin D$, $\sin C-\sin D$ , $\sin 2\theta $ for simplifying the term on the left-hand side. After that, we will easily prove the desired result.

Complete step-by-step answer:
Given:
We have to prove the following equation:
${{\sin }^{2}}6x-{{\sin }^{2}}4x=\sin 2x\sin 10x$
Now, we will simplify the term on the left-hand side and prove that it is equal to the term on the right-hand side.
Now, before we proceed we should know the following formulas:
$\begin{align}
  & {{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)........................\left( 1 \right) \\
 & \sin C+\sin D=2\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)...................\left( 2 \right) \\
 & \sin C-\sin D=2\cos \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)...................\left( 3 \right) \\
 & 2\sin \theta \cos \theta =\sin 2\theta .....................................................\left( 4 \right) \\
\end{align}$
Now, we will use the above four formulas to simplify the term on the left-hand side.
On the left-hand side, we have ${{\sin }^{2}}6x-{{\sin }^{2}}4x$ .
Now, we will use the formula from the equation (1) to write ${{\sin }^{2}}6x-{{\sin }^{2}}4x=\left( \sin 6x+\sin 4x \right)\left( \sin 6x-\sin 4x \right)$ in the term on the left-hand side. Then,
$\begin{align}
  & {{\sin }^{2}}6x-{{\sin }^{2}}4x \\
 & \Rightarrow \left( \sin 6x+\sin 4x \right)\left( \sin 6x-\sin 4x \right) \\
\end{align}$
Now, we will use the formula from the equation (2) to write $\sin 6x+\sin 4x=2\sin 5x\cos x$ and the formula from the equation (3) to write $\sin 6x-\sin 4x=2\cos 5x\sin x$ in the above expression. Then,
$\begin{align}
  & \left( \sin 6x+\sin 4x \right)\left( \sin 6x-\sin 4x \right) \\
 & \Rightarrow \left( 2\sin \left( \dfrac{6x+4x}{2} \right)\cos \left( \dfrac{6x-4x}{2} \right) \right)\left( 2\cos \left( \dfrac{6x+4x}{2} \right)\sin \left( \dfrac{6x-4x}{2} \right) \right) \\
 & \Rightarrow \left( 2\sin 5x\cos x \right)\left( 2\cos 5x\sin x \right) \\
 & \Rightarrow \left( 2\sin 5x\cos 5x \right)\left( 2\sin x\cos x \right) \\
\end{align}$
Now, we will use the formula from the equation (4) to write $2\sin 5x\cos 5x=\sin 10x$ and $2\sin x\cos x=\sin 2x$ in the above expression. Then,
$\begin{align}
  & \left( 2\sin 5x\cos 5x \right)\left( 2\sin x\cos x \right) \\
 & \Rightarrow \left( \sin 10x \right)\times \left( \sin 2x \right) \\
 & \Rightarrow \sin 2x\sin 10x \\
\end{align}$
Now, from the above result, we conclude that the value of the expression ${{\sin }^{2}}6x-{{\sin }^{2}}4x$ will be equal to the value of the expression $\sin 2x\sin 10x$ . Then,
${{\sin }^{2}}6x-{{\sin }^{2}}4x=\sin 2x\sin 10x$
Now, from the above result, we conclude that the term on the left-hand side is equal to the term on the right-hand side.
Thus, ${{\sin }^{2}}6x-{{\sin }^{2}}4x=\sin 2x\sin 10x$ .
Hence, proved.

Note: Here, the student should first understand what we have to prove in the question. After that, we should proceed in a stepwise manner and apply trigonometric formulas of $\sin C+\sin D$ and $\sin C-\sin D$ correctly. Moreover, while simplifying we should be aware of the result and avoid calculation mistakes while solving.