# What is the equation of the normal line of f(x)= sinx-cos2x at x = pi/8?

Mar 1, 2017

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

#### Explanation:

Find the first derivative $\to$the slope function:
$f ' \left(x\right) = \cos x - \left(- \sin 2 x\right) \left(2\right) = \cos x + 2 \sin 2 x$

Find the slope of the tangent line: ${m}_{t} = f ' \left(\frac{\pi}{8}\right) = \cos \left(\frac{\pi}{8}\right) + 2 \sin \left(\frac{\pi}{4}\right) = \frac{\sqrt{2 + \sqrt{2}}}{2} + \frac{2 \sqrt{2}}{2}$
${m}_{t} = \frac{\sqrt{2 + \sqrt{2}} + 2 \sqrt{2}}{2}$

The normal line is the perpendicular line which the negative reciprocal of the tangent line for its slope:
${m}_{n} = - \frac{2}{\sqrt{2 + \sqrt{2}} + 2 \sqrt{2}}$

${y}_{n} - {y}_{1} = {m}_{n} \left(x - {x}_{1}\right)$

$f \left(\frac{\pi}{8}\right) = \sin \left(\frac{\pi}{8}\right) - \cos \left(\frac{\pi}{4}\right) = \frac{\sqrt{2 - \sqrt{2}}}{2} - \frac{\sqrt{2}}{2} = \frac{\left(\sqrt{2 - \sqrt{2}}\right) - \sqrt{2}}{2}$

${y}_{n} - {y}_{1} = {m}_{n} \left(x - {x}_{1}\right) :$

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

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