Question

What is the equation of the tangent line of `f(x)=4secx–8cosx` at `x=(pi/3)`?

Answer 1

`y-4=12sqrt3(x-pi/3)`

Explanation:

Find the point the tangent line will intercept.

`f(pi/3)=4sec(pi/3)-8cos(pi/3)=4(2)-8(1/2)=4`

The tangent line will intercept the point `(pi/3,4)`.

In order to find the slope of the tangent line, we must find the derivative of the function. In order to do this, the two following trigonometric identities should be known:

• `d/dx[secx]=secxtanx`
• `d/dx[cosx]=-sinx`

The derivative is

`f(x)=4secx-8cosx`

`f'(x)=4secxtanx+8sinx`

The slope of the tangent line at `x=pi/3` can be found through evaluating `f'(pi/3)`.

`f'(pi/3)=4sec(pi/3)tan(pi/3)+8sin(pi/3)=4(2)(sqrt3)+8(sqrt3/2)=12sqrt3`

The tangent line will intercept the point `(pi/3,4)` and have a slope of `12sqrt3`.

These can be related in a linear equation in point-slope form, which uses a point `(x_1,y_1)` and slope `m`:

`y-y_1=m(x-x_1)`

Giving the equation of the tangent line

`y-4=12sqrt3(x-pi/3)`

Graphed are the original function and the tangent line (zoomed in):

graph{(4secx-8cosx-y)(y-4-12sqrt3(x-pi/3))=0 [-1.5, 2, -6, 15]}