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

Question

Ezekiel Alexander · Accepted Answer

#d/(dx)(sinx/(1-cosx))=-cscxcotx-csc^2x#

and not #-cscx# #sinx/(1-cosx)# = #(sinx(1+cosx))/((1-cosx)(1+cosx))# = #(sinx(1+cosx))/(1-cos^2x)# = #(sinx(1+cosx))/sin^2x# = #(1+cosx)/sinx# = #cscx+cotx# As such #d/(dx)(sinx/(1-cosx))# = #d/(dx)(cscx+cotx)# = #d/(dx)cscx+d/(dx)cotx# = #-cscxcotx-csc^2x#

Paisley Johnson · Answer

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

Let's use the product rule.

#d/(dx)[(sinx)/(1-cosx)]#

#= sinx * -1/(1-cosx)^2 * -(-sinx) + 1/(1-cosx) * cosx#

#= -(sin^2x)/(1-cosx)^2 + cosx/(1-cosx)#

#= -(sin^2x)/(1-cosx)^2 + (cosx(1-cosx))/(1-cosx)^2#

#= -(sin^2x)/(1-cosx)^2 + (cosx - cos^2x)/(1-cosx)^2#

#= (cosx - cos^2x - sin^2x)/(1-cosx)^2#

#= (cosx - (cos^2x + sin^2x))/(1-cosx)^2#

#= (cosx - 1)/(1-cosx)^2#

#= -cancel(1 - cosx)/(1-cosx)^cancel(2)#

#= -1/(1-cosx)#

But we know that #-1/sinx = -cscx#. Therefore, we claim that

#-1/(1-cosx) = -1/sinx#

Therefore:

#sinx/(1-cosx) = 1#

#sinx = 1 - cosx#

#sinx + cosx ne 1#

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

#sinx + cosx#:

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 #RR#.

Aurora Alvarado · Answer

undefined To find the derivative of $\frac{\sin(x)}{1 - \cos(x)}$, use the quotient rule:
$$ \frac{d}{dx}\left(\frac{\sin(x)}{1 - \cos(x)}\right) = \frac{(1 - \cos(x))\cos(x) - (-\sin(x))(\sin(x))}{(1 - \cos(x))^2} $$
$$ = \frac{\cos(x) - \cos^2(x) + \sin^2(x)}{(1 - \cos(x))^2} $$
$$ = \frac{\cos(x) - \cos^2(x) + (1 - \cos^2(x))}{(1 - \cos(x))^2} $$
$$ = \frac{\cos(x) - \cos^2(x) + 1 - \cos^2(x)}{(1 - \cos(x))^2} $$
$$ = \frac{1 - \cos^2(x) - \cos^2(x) + \cos(x) + 1}{(1 - \cos(x))^2} $$
$$ = \frac{2 - 2\cos^2(x) + \cos(x)}{(1 - \cos(x))^2} $$
$$ = \frac{2(1 - \cos^2(x)) + \cos(x)}{(1 - \cos(x))^2} $$
$$ = \frac{2\sin^2(x) + \cos(x)}{(1 - \cos(x))^2} $$
$$ = \frac{2\sin^2(x) + \cos(x)}{\sin^2(x)} $$
$$ = \frac{2\sin^2(x)}{\sin^2(x)} + \frac{\cos(x)}{\sin^2(x)} $$
$$ = 2 + \frac{\cos(x)}{\sin^2(x)} $$
$$ = 2 + \csc^2(x) $$
$$ = -\csc^2(x) $$
Therefore, the derivative of $\frac{\sin(x)}{1 - \cos(x)}$ is $-\csc^2(x)$.