Question

Prove that $\dfrac{{\sin x - \sin 3x}}{{{{\sin }^2}x - {{\cos }^2}x}}$ is equal to $2\sin x$.

Answer 1

Hint: We will write sin3x in terms of sin function and also write the denominator in terms of sin. We can write $\sin 3x = 3\sin x - 4{\sin ^3}x$ . After substituting the value, we will try to take some common terms. And cancel the common parts which are common in both numerator and denominator.

Complete step-by-step solution:
We have to prove $\dfrac{{\sin x - \sin 3x}}{{{{\sin }^2}x - {{\cos }^2}x}} = 2\sin x$
Our first approach to solve these types of questions is that we will try to convert all of the given terms in terms of the result as in the given question, we will convert all our terms in terms of sin.
We know that the formula of $\sin 3x = 3\sin x - 4{\sin ^3}x$ (1)
And we also know that ${\sin ^2}x + {\cos ^2}x = 1$ (2)
We will subtract ${\sin ^2}x$ on both side of equation 2, we get
${\cos ^2}x = 1 - {\sin ^2}x$ (3)
Now, we will substitute the value of sin3x and ${\cos ^2}x$ in the equation
$ = \dfrac{{\sin x - \sin 3x}}{{{{\sin }^2}x - {{\cos }^2}x}}$
We have substituted the values
$ = \dfrac{{\sin x - 3\sin x + 4{{\sin }^3}x}}{{{{\sin }^2}x - 1 + {{\sin }^2}x}}$
$ = \dfrac{{4{{\sin }^3}x - 2\sin x}}{{2{{\sin }^2}x - 1}}$
We take 2sinx as common from the numerator
$ = \dfrac{{2\sin x\left( {2{{\sin }^2}x - 1} \right)}}{{2{{\sin }^2}x - 1}}$
We divide the numerator and denominator by $2{\sin ^2}x - 1$ , we get
$ = 2\sin x$
Hence, we proved that $\dfrac{{\sin x - \sin 3x}}{{{{\sin }^2}x - {{\cos }^2}x}} = 2\sin x$.

Note: We can also solve this question by converting the numerator in terms of \[\sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)\] where A is x and B is 3x. we will convert the denominator in terms of cos2x. Then using trigonometric functions, we will simplify them.