Question

How do you prove ${\sin }^{2}x-{\sin }^{2}y=\sin \left(x+y\right)\sin \left(x-y\right)$?

Answer 1

Proof of given relation:

${\sin }^{2}x-{\sin }^{2}y=\sin \left(x+y\right)\sin \left(x-y\right)$

Here we have $RHS=\sin \left(x+y\right)\sin \left(x-y\right)$

Therefore,

$\sin \left(x+y\right)\sin \left(x-y\right)=\left(\sin x\cos y+\cos x\sin y\right)\left(\sin x\cos y-\cos x\sin y\right)$

$$\begin{gathered} \left[\because \sin \left(a+b\right)=\mathrm{sina}\mathrm{cosb}+\mathrm{cosa}\mathrm{sinb}\right] \\ \left[\because \sin \left(a-b\right)=\mathrm{sina}\mathrm{cosb}-\mathrm{cosa}\mathrm{sinb}\right] \end{gathered}$$

$={\sin }^{2}x{\cos }^{2}y-\sin x\sin y\cos x\cos y+\sin x\sin y\cos x\cos y-{\cos }^{2}x{\sin }^{2}y$

$={\sin }^{2}x{\cos }^{2}y-{\cos }^{2}x{\sin }^{2}y$

$={\sin }^{2}x\left(1-{\sin }^{2}y\right)-\left(1-{\sin }^{2}x\right){\sin }^{2}y$ $\left[\because si{n}^{2}x+co{s}^{2}x=1\right]$

$={\sin }^{2}x-{\sin }^{2}x{\sin }^{2}y-{\sin }^{2}y+{\sin }^{2}x{\sin }^{2}y$

$={\sin }^{2}x-{\sin }^{2}y$

$=LHS$

Hence, $\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{x}\mathbf{-}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{y}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\left(x+y\right)\mathit{s}\mathit{i}\mathit{n}\left(x-y\right)$ is proved.