Show that the derivative of (sinx)/(1 - cosx) is -cscx?

Mar 12, 2017

$\frac{d}{\mathrm{dx}} \left(\sin \frac{x}{1 - \cos x}\right) = - \csc x \cot x - {\csc}^{2} x$

and not $- \csc x$

Explanation:

$\sin \frac{x}{1 - \cos x}$

= $\frac{\sin x \left(1 + \cos x\right)}{\left(1 - \cos x\right) \left(1 + \cos x\right)}$

= $\frac{\sin x \left(1 + \cos x\right)}{1 - {\cos}^{2} x}$

= $\frac{\sin x \left(1 + \cos x\right)}{\sin} ^ 2 x$

= $\frac{1 + \cos x}{\sin} x$

= $\csc x + \cot x$

As such $\frac{d}{\mathrm{dx}} \left(\sin \frac{x}{1 - \cos x}\right)$

= $\frac{d}{\mathrm{dx}} \left(\csc x + \cot x\right)$

= $\frac{d}{\mathrm{dx}} \csc x + \frac{d}{\mathrm{dx}} \cot x$

= $- \csc x \cot x - {\csc}^{2} x$

Mar 12, 2017

We can't, because it's false...

Let's use the product rule.

$\frac{d}{\mathrm{dx}} \left[\frac{\sin x}{1 - \cos x}\right]$

$= \sin x \cdot - \frac{1}{1 - \cos x} ^ 2 \cdot - \left(- \sin x\right) + \frac{1}{1 - \cos x} \cdot \cos x$

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

$= - \frac{{\sin}^{2} x}{1 - \cos x} ^ 2 + \frac{\cos x \left(1 - \cos x\right)}{1 - \cos x} ^ 2$

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

$= \frac{\cos x - {\cos}^{2} x - {\sin}^{2} x}{1 - \cos x} ^ 2$

$= \frac{\cos x - \left({\cos}^{2} x + {\sin}^{2} x\right)}{1 - \cos x} ^ 2$

$= \frac{\cos x - 1}{1 - \cos x} ^ 2$

$= - \frac{\cancel{1 - \cos x}}{1 - \cos x} ^ \cancel{2}$

$= - \frac{1}{1 - \cos x}$

But we know that $- \frac{1}{\sin} x = - \csc x$. Therefore, we claim that

$- \frac{1}{1 - \cos x} = - \frac{1}{\sin} x$

Therefore:

$\sin \frac{x}{1 - \cos x} = 1$

$\sin x = 1 - \cos x$

$\sin x + \cos x \ne 1$

Clearly, we have found that this identity is false. More definitively, we can plot this equation to get:

$\sin x + \cos x$:

graph{sinx + cosx [-10, 10, -5, 5]}

As this equation is not a straight horizontal line, it cannot be equal to $1$ for all $x$ in $\mathbb{R}$.