# Thread: differentiate the trig function

1. ## differentiate the trig function

1. sinx/1-cosx.
(1-cosx)(sinx)'-(sinx)(1-cosx)'
-----------------------------------
(1-cosx)^2

(1-cosx)(cosx)-(sinx)(sinx)
--------------------------------
(1-cosx)^2

(cosx-cosx^2)-sinx^2
--------------------------------
(1-cosx)^2

the answer comes out to be 1/cos(x)-1.

2. 1+sinx/2-cosx
(2-cosx)(1+sinx)'-(1+sinx)(2-cosx)'
---------------------------------------
(2-cosx)^2

(2-cosx)(cosx)-(1+sinx)(sinx)
---------------------------------------
(2-cosx)^2

2cosx-cosx^2-sinx-sinx^2
---------------------------------------
(2-cosx)^2

the answer comes out to be (2cosx -sinx -1)/(2-cosx)^2.

3.sinx+cosx/sinx-cosx

(sinx-cosx)(sinx+cosx)'-(sinx+cosx)(sinx-cosx)'
-----------------------------------------------
(sinx-cosx)^2

(sinx-cosx)(cos-sinx)-(sinx+cosx)(cosx+sinx)
-----------------------------------------------
(sinx-cosx)^2

and after a whole lot of crossing out, I got (sinxcosx+sinxcosx-sinxcosx+sinxcosx)/(sinx-cosx)^2

but the answer says -2/(sinx-cosx)^2

2. See here.

3. 1.)$\displaystyle \frac{\sin{x}}{1-\cos{x}}$
$\displaystyle =\frac{(1-\cos{x})(\sin{x})' - (1-\cos{x})'(\sin{x})}{(1-\cos{x})^2}$
$\displaystyle =\frac{(1-\cos{x})(\cos{x}) - (-(-\sin{x})(\sin{x}))}{(1-\cos{x})^2}$
$\displaystyle =\frac{\cos{x} - \cos^2{x} - \sin^2{x}}{(1-\cos{x})^2}$
$\displaystyle =\frac{\cos{x} - (\cos^2{x} + \sin^2{x})}{(1-\cos{x})^2}$
$\displaystyle =\frac{\cos{x} - 1}{(1-\cos{x})^2}$
$\displaystyle =-\frac{1-\cos{x}}{(1-\cos{x})^2}$
cancel 1- cos(x)
$\displaystyle =-\frac{1}{1-\cos{x}}$

yep there's a negetive sign

4. Originally Posted by driver327
2. 1+sinx/2-cosx
(2-cosx)(1+sinx)'-(1+sinx)(2-cosx)'
---------------------------------------
(2-cosx)^2

(2-cosx)(cosx)-(1+sinx)(sinx)
---------------------------------------
(2-cosx)^2

2cosx-cosx^2-sinx-sinx^2
---------------------------------------
(2-cosx)^2

the answer comes out to be (2cosx -sinx -1)/(2-cosx)^2.
you are correct you just need to use some identities:
$\displaystyle \frac{2\cos{x}-\sin{x}-(\cos^2{x}+\sin^2{x})}{(2-\cos{x})^2}$

where $\displaystyle \cos^2{x}+\sin^2{x}$ is 1

5. Hello, driver327!

$\displaystyle 3)\;\;y \:=\:\frac{\sin x + \cos x}{\sin x - \cos x}$

$\displaystyle \frac{dy}{dx} \;=\;\frac{(\sin x - \cos x)(\cos x-\sin x) -(\sin x + \cos x)(\cos x +\sin x)}{(\sin x-\cos x)^2}$ . . . . Keep going

Multiply: .$\displaystyle \frac{(\sin x\cos x - \sin^2x - \cos^2x+ \sin x \cos x) - (\sin x\cos x + \sin^2x + \cos^2x + \sin x\cos x)}{(\sin x-\cos x)^2}$

. . $\displaystyle = \;\frac{\left[2\sin x\cos x - \overbrace{(\sin^2x + \cos^2x)}^{\text{This is 1}}\right] - \left[2\sin x\cos x + \overbrace{(\sin^2x + \cos^2x)}^{\text{This is 1}}\right]} {(\sin x -\cos x)^2}$

. . $\displaystyle = \;\frac{2\sin x\cos x - 1 - 2\sin x\cos x - 1}{(\sin x-\cos x)^2}$

. . $\displaystyle = \;\frac{-2}{(\sin x -\cos x)^2}$