Question

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

Answer 1

`f(x)= (-2x) /(sqrt(3)+2)-(1+sqrt(3))/2+(8pi) /(3sqrt(3)+6)`

Explanation:

The normal line is perpendicular to the slope at that point, so we have to derive the formula, plug in `x=(4pi)/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=(4pi)/3` to get the slope at this point is `((sqrt(3)+2)/2)`,but to find the perpendicular slope you use the negative reciprocal of that slope, which is `-2/(sqrt(3)+2)`.

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