# How do you find the tangent line approximation to f(x)=cos(x) at x=pi/4 ?

Sep 24, 2014

In this problem we need to follow 5 steps.

1) Find the value of y given the x-value of $\frac{\pi}{4}$. This gives you the point of tangency.

2) Find the derivative of $f \left(x\right)$

3) Substitute in the x-value of $\frac{\pi}{4}$ to find the numeric slope.

4) Substitute the slope and the $x$ and $y$ values to find the y-intercept.

5) Use the y-intercept and slope to find the equation of the of the tangent line.

$f \left(x\right) = \cos \left(x\right)$

$f \left(\frac{\pi}{4}\right) = \cos \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$

Point of tangency: $\left(\frac{\pi}{4} , \frac{\sqrt{2}}{2}\right)$

$f ' \left(x\right) = - \sin \left(x\right)$

$f ' \left(\frac{\pi}{4}\right) = - \sin \left(\frac{\pi}{4}\right) = - \frac{\sqrt{2}}{2} \implies s l o p e \mathmr{and} m$

$y = m x + b$

$\frac{\sqrt{2}}{2} = \left(- \frac{\sqrt{2}}{2}\right) \left(\frac{\pi}{4}\right) + b$

$\frac{\sqrt{2}}{2} = \left(\frac{- \sqrt{2} \pi}{8}\right) + b$

$\left(\frac{\sqrt{2} \pi}{8}\right) + \frac{\sqrt{2}}{2} = b$

$1.2625 = b$

$y = m x + b$

$y = \frac{- \sqrt{2}}{2} x + 1.2625 \implies$Equation of the tangent line