# Thread: Equation of a line tangent

1. ## 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

2. 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:

$\displaystyle \frac{d}{dx}\left[\ln{(\sin{x})}\right]=\cot{x}\Rightarrow\cot{x}=1\Rightarrow\cos{x}=\si n{x}\Rightarrow{x}=\frac{\pi}{4}$

3. Originally Posted by VonNemo19
Take the derivative:

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

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

4. Originally Posted by skeeter
correction ...

$\displaystyle x = \frac{\pi}{4}$
Thank you. Got a little ahead of myself...