# How do you verify the identity: -cotx =(sin3x+sinx)/(cos3x-cosx)?

##### 1 Answer
Feb 6, 2015

There are some formulas, named sum-to-product, that say:

$\sin \theta + \sin \phi = 2 \sin \left(\frac{\theta + \phi}{2}\right) \cos \left(\frac{\theta - \phi}{2}\right)$,
$\sin \theta - \sin \phi = 2 \cos \left(\frac{\theta + \phi}{2}\right) \sin \left(\frac{\theta - \phi}{2}\right)$,
$\cos \theta + \sin \phi = 2 \cos \left(\frac{\theta + \phi}{2}\right) \cos \left(\frac{\theta - \phi}{2}\right)$,
$\cos \theta + \cos \phi = - 2 \sin \left(\frac{\theta + \phi}{2}\right) \sin \left(\frac{\theta - \phi}{2}\right)$.

So:

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

$- \cot x = - \frac{\cos x}{\sin} x$.

That's all.
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