# How do you verify -cot(x/2) = (sin2x + sinx) / (cos2x - cosx)?

Nov 2, 2015

Verify trig expression
$- \cot \left(\frac{x}{2}\right) = \frac{\sin 2 x + \sin x}{\cos 2 x - \cos x}$

#### Explanation:

Use the trig identities to transform the right side:
sin 2x = sin x.cos x
$\cos 2 x = 2 {\cos}^{2} x - 1$.
Transform the right side:
RS = (sin 2x + sin x)/(cos 2x - cos x) = (sin x(2cos x + 1))/(2cos^2 x - cos x - 1 =

Factor the trinomial $\left(2 {\cos}^{2} x - \cos x - 1\right) .$
Since (a + b + c = 0), use the Shortcut, the 2 factors are (cos x - 1) and (2cos x + 1). Finally,

$R S = \frac{\sin x \left(2 \cos x + 1\right)}{\left(\cos x - 1\right) \left(2 \cos x + 1\right)} = \frac{\sin x}{\cos x - 1}$
Since:
$\sin x = 2 \sin \left(\frac{x}{2}\right) . \cos \left(\frac{x}{2}\right) \mathmr{and}$
$\left(\cos x - 1\right) = - 2 {\sin}^{2} \left(\frac{x}{2}\right)$, therefor
$R S = \frac{2 \sin \left(\frac{x}{2}\right) \left(\cos \frac{x}{2}\right)}{- 2 {\sin}^{2} \left(\frac{x}{2}\right)} =$
$= - \cos \frac{\frac{x}{2}}{\sin \left(\frac{x}{2}\right)} = - \cot \frac{x}{2}$