# Question #73fd2

Apr 27, 2017

See proof below

#### Explanation:

We need

$\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$

${\sin}^{2} x + {\cos}^{2} x = 1$

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

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

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

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

Therefore,

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

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

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

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

So,

$L H S = \cos \left(x + y\right) \cdot \cos \left(x - y\right) + \sin \left(x + y\right) \cdot \sin \left(x - y\right)$

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

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

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

$= L H S$

$Q E D$