Question

Prove the following:
$\dfrac{\sin x-\sin 3x}{{{\sin }^{2}}x-{{\cos }^{2}}x}=2\sin x$

Answer 1

Hint:If you look carefully, the numerator of the L.H.S expression is in the form of $\sin C-\sin D$ and the denominator of the L.H.S expression is in the form of ${{\cos }^{2}}\theta -{{\sin }^{2}}\theta $. So, apply the identity of$\sin C-\sin D$ and ${{\cos }^{2}}\theta -{{\sin }^{2}}\theta $ in the numerator and denominator of L.H.S respectively and then solve.

Complete step-by-step answer:
We have to prove the following expression:
$\dfrac{\sin x-\sin 3x}{{{\sin }^{2}}x-{{\cos }^{2}}x}=2\sin x$
Solving L.H.S of the expression,
$\dfrac{\sin x-\sin 3x}{{{\sin }^{2}}x-{{\cos }^{2}}x}$
The numerator of the above expression is in the form of $\sin C-\sin D$ and we know that:
$\sin C-\sin D=2\cos \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)$
Applying this formula in $\sin x-\sin 3x$ we get,
$\sin x-\sin 3x=2\cos 2x\sin (-x)$
Now, the denominator of the L.H.S expression is in the form of ${{\cos }^{2}}\theta -{{\sin }^{2}}\theta $ and we know that:
${{\cos }^{2}}\theta -{{\sin }^{2}}\theta =\cos 2\theta $
Applying this formula in ${{\sin }^{2}}x-{{\cos }^{2}}x$ we get,
${{\sin }^{2}}x-{{\cos }^{2}}x=-\cos 2x$
Substituting the value of $\sin x-\sin 3x$ and ${{\sin }^{2}}x-{{\cos }^{2}}x$ in the L.H.S expression of the given question we get,
$\dfrac{\sin x-\sin 3x}{{{\sin }^{2}}x-{{\cos }^{2}}x}$
$=\dfrac{2\cos 2x\sin (-x)}{-\cos 2x}$
From the trigonometric properties we know that:
$\sin (-x)=-\sin x$.
Now, we are going to substitute this value in the above expression and $\cos 2x$ will be cancelled out from the numerator and the denominator of the above expression.
$\begin{align}
  & \dfrac{-2\sin x}{-1} \\
 & =2\sin x \\
\end{align}$
The answer we get from solving the L.H.S of the given expression is $2\sin x$ which is equal to R.H.S.
Hence, we have proved that L.H.S = R.H.S of the given expression.

Note: You can do the above proof by cross – multiplying the given equation:
$\dfrac{\sin x-\sin 3x}{{{\sin }^{2}}x-{{\cos }^{2}}x}=2\sin x$
$\sin x-\sin 3x=2\sin x\left( {{\sin }^{2}}x-{{\cos }^{2}}x \right)$
Multiplying -1 on both the sides we get,
$\sin 3x-\sin x=2\sin x\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right)$…………. Eq. (1)
Solving R.H.S of the above equation:
$2\sin x\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right)$
We know that ${{\cos }^{2}}x-{{\sin }^{2}}x=\cos 2x$ so using this relation in the above expression.
$2\sin x\cos 2x$
Now, solving R.H.S of the eq. (1) we get,
$\sin 3x-\sin x$
Using the formula of $\sin C-\sin D$ in the above equation we get,
$\sin 3x-\sin x=2\cos 2x\sin (x)$
The above simplification of L.H.S of eq. (1) has given the value $2\cos 2x\sin x$ and R.H.S we have solved above has given the value $2\sin x\cos 2x$.
As you can see L.H.S = R.H.S. so we have proved.