# What is the equation of the line that is normal to f(x)= cosx-sin2x at  x=(4pi)/3 ?

Feb 11, 2018

$f \left(x\right) = \frac{- 2 x}{\sqrt{3} + 2} - \frac{1 + \sqrt{3}}{2} + \frac{8 \pi}{3 \sqrt{3} + 6}$

#### Explanation:

The normal line is perpendicular to the slope at that point, so we have to derive the formula, plug in $x = \frac{4 \pi}{3}$ to find the slope at that point, find the negative reciprocal of the point to get the perpendicular slope, then use y=mx+b to find the y-intercept of the line

f'(x)=d/dx[cos(x)-sin(2x)]=-sin(x)-2cos(x)]
Plug in $x = \frac{4 \pi}{3}$ to get the slope at this point is $\left(\frac{\sqrt{3} + 2}{2}\right)$,but to find the perpendicular slope you use the negative reciprocal of that slope, which is $- \frac{2}{\sqrt{3} + 2}$.

Now with the slope and the point it must intersect $\left(\frac{4 \pi}{3} , - \frac{1 + \sqrt{3}}{2}\right)$, find the y-intercept by using y=mx=b to get that messy equation in the answer box.