# What is the equation of the line normal to  f(x)=cos(5x+pi/4) at  x=pi/3?

color(red)(y-((sqrt2+sqrt6))/4=-((sqrt2+sqrt6))/5*(x-pi/3)

#### Explanation:

Given $f \left(x\right) = \cos \left(5 x + \frac{\pi}{4}\right)$ at ${x}_{1} = \frac{\pi}{3}$

Solve for the point $\left({x}_{1} , {y}_{1}\right)$

$f \left(\frac{\pi}{3}\right) = \cos \left(\frac{5 \cdot \pi}{3} + \frac{\pi}{4}\right) = \frac{\sqrt{2} + \sqrt{6}}{4}$

point $\left({x}_{1} , {y}_{1}\right) = \left(\frac{\pi}{3} , \frac{\sqrt{2} + \sqrt{6}}{4}\right)$

Solve for the slope m

$f ' \left(x\right) = - 5 \cdot \sin \left(5 x + \frac{\pi}{4}\right)$

$m = - 5 \cdot \sin \left(\frac{5 \pi}{3} + \frac{\pi}{4}\right)$

$m = \frac{- 5 \left(\sqrt{2} - \sqrt{6}\right)}{4}$

for the normal line ${m}_{n}$

${m}_{n} = - \frac{1}{m} = - \frac{1}{\frac{- 5 \left(\sqrt{2} - \sqrt{6}\right)}{4}} = \frac{4}{5 \left(\sqrt{2} - \sqrt{6}\right)}$
${m}_{n} = - \frac{\sqrt{2} + \sqrt{6}}{5}$

Solve the normal line

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

color(red)(y-((sqrt2+sqrt6))/4=-((sqrt2+sqrt6))/5*(x-pi/3)

Kindly see the graph of $y = \cos \left(5 x + \frac{\pi}{4}\right)$ and the normal line $y - \frac{\left(\sqrt{2} + \sqrt{6}\right)}{4} = - \frac{\left(\sqrt{2} + \sqrt{6}\right)}{5} \cdot \left(x - \frac{\pi}{3}\right)$

graph{(y-cos (5x+pi/4))(y-((sqrt2+sqrt6))/4+((sqrt2+sqrt6))/5*(x-pi/3))=0[-5,5,-2.5,2.5]}

God bless....I hope the explanation is useful.