# What is the equation of the normal line of f(x)= cosx at x = pi/8?

Feb 7, 2016

$y - \frac{\pi}{8} = \sqrt{4 + 2 \sqrt{2}} \left(x - \frac{\sqrt{2 + \sqrt{2}}}{2}\right)$

#### Explanation:

We can find the point the normal line will intercept:

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

The normal line passes through the point $\left(\frac{\pi}{8} , \frac{\sqrt{2 + \sqrt{2}}}{2}\right)$
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The exact value can be found through the cosine half-angle formula:

$\cos \left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \cos \theta}{2}}$

To find the slope of the tangent line, find $f ' \left(\frac{\pi}{8}\right)$. First, find $f ' \left(x\right)$.

The derivative of cosine is negative sine:

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

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

The slope of the tangent line is

$f ' \left(\frac{\pi}{8}\right) = - \sin \left(\frac{\pi}{8}\right) = - \frac{\sqrt{2 - \sqrt{2}}}{2}$

This can be found through the sine half-angle formula:

$\sin \left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \cos \theta}{2}}$

Now, recall that we are looking for the normal line, which is perpendicular to the tangent line. Perpendicular lines have opposite reciprocal slopes:

• Slope of tangent line: $\text{ } - \frac{\sqrt{2 - \sqrt{2}}}{2}$
• Slope of normal line: $\text{ } \frac{2}{\sqrt{2 - \sqrt{2}}} = \sqrt{4 + 2 \sqrt{2}}$

For an explanation of how to simplify the slope of normal line, ask, but I'll focus more on the calculus here.

We now know that the normal line passes through the point $\left(\frac{\pi}{8} , \frac{\sqrt{2 + \sqrt{2}}}{2}\right)$ and has a slope of $\sqrt{4 + 2 \sqrt{2}}$.

The equation of such line in point-slope form is:

$y - \frac{\pi}{8} = \sqrt{4 + 2 \sqrt{2}} \left(x - \frac{\sqrt{2 + \sqrt{2}}}{2}\right)$