# Equation of a line tangent

• Jul 26th 2009, 04:48 PM
05holtel
Equation of a line tangent
What is the value of the constant c so that x-y=c is the equation of the line tangent to the graph f(x)=In(Sin(x)) where 0<x<pi

I know that line x-y=c has a slope of 1, so therefore f'x = 1
I do not what to do next
• Jul 26th 2009, 05:15 PM
VonNemo19
Quote:

Originally Posted by 05holtel
What is the value of the constant c so that x-y=c is the equation of the line tangent to the graph f(x)=In(Sin(x)) where 0<x<pi

I know that line x-y=c has a slope of 1, so therefore f'x = 1
I do not what to do next

Take the derivative:

$\frac{d}{dx}\left[\ln{(\sin{x})}\right]=\cot{x}\Rightarrow\cot{x}=1\Rightarrow\cos{x}=\si n{x}\Rightarrow{x}=\frac{\pi}{4}$
• Jul 26th 2009, 05:24 PM
skeeter
Quote:

Originally Posted by VonNemo19
Take the derivative:

$\frac{d}{dx}\left[\ln{(\sin{x})}\right]=\cot{x}\Rightarrow\cot{x}=1\Rightarrow\cos{x}=\si n{x}\Rightarrow{x}=\frac{\sqrt{2}}{2}$

correction ...

$x = \frac{\pi}{4}$
• Jul 26th 2009, 05:25 PM
VonNemo19
Quote:

Originally Posted by skeeter
correction ...

$x = \frac{\pi}{4}$

Thank you. Got a little ahead of myself...