# Question ffc58

Aug 27, 2017

$\text{see explanation}$

#### Explanation:

$\text{using the "color(blue)"trigonometric identities}$

•color(white)(x)cos(A+B)=cosAcosB-sinAsinB

•color(white)(x)sin(A-B)=sinAcosB-cosAsinB

•color(white)(x)cos2A=cos^2A-sin^2A#

$\text{consider left hand side}$

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

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

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

$= \sin y \cos 2 x = \text{ right hand side "rArr" proved}$

Aug 27, 2017

We have: $\sin \left(x\right) \cos \left(x + y\right) - \cos \left(x\right) \sin \left(x - y\right)$

Let's apply the compound angle identities for $\sin \left(x\right)$ and $\cos \left(x\right)$:

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

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

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

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

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

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

Then, let's apply the double angle identity for $\cos \left(x\right)$; $\cos \left(2 x\right) = {\cos}^{2} \left(x\right) - {\sin}^{2} \left(x\right)$:

$= - \sin \left(y\right) \cdot \cos \left(2 x\right)$

$= - \cos \left(2 x\right) \sin \left(y\right)$