Question

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

Answer 1

We have

L.H.S. = $\frac{\text{sin x - sin 3x}}{{\text{sin}}^{2}x-{\text{cos}}^{2}x}$

= $\frac{-\left(\text{sin 3x - sin x}\right)}{-({\text{cos}}^{2}x-{\text{sin}}^{2}x)}$

= $\frac{\text{2 cos}\left(\frac{3x+x}{2}\right)\text{sin}\left(\frac{3x-x}{2}\right)}{\text{cos 2x}}$

$$\left[\begin{array}{c}\because \text{sin C -sin D = 2 cos}\left(\frac{C+D}{2}\right)\text{sin}\left(\frac{C-D}{2}\right)\\ {\text{cos 2A = cos}}^{2}A-{\text{sin}}^{2}A\end{array}\right]$$

= $\frac{\text{2 cos 2x sin x}}{\text{cos 2x}}$ = 2 sin x

= R.H.S.