Proving Trigonometric Identities?

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

Jun 21, 2018

See below

Explanation:

Use the following identities:

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

The expression becomes

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

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

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

And since ${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$, the result is proven