Question

What is the equation of the line tangent to `f(x)=secx - 2cosx` at `x=pi/3`?

Answer 1

Calculate the derivative (slope of the line), and find `f(pi/3)` so you have a point for point slope form to find the line equation.

`y = (3sqrt3)x + 1 - (sqrt3)pi`

Explanation:

To find the equation of the line in question, we need both a point, and a slope. The point will be found by simply finding `f(pi/3)`

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

We know, either from use of a calculator or from learning the values of the trig functions for angles such as `npi, npi/2, npi/3, npi/4, npi/6`, that `cos(pi/3) = 1/2`. Thus...

`f(pi/3) = 1/(1/2) - 2(1/2) = 2-1 = 1`

Thus the point `(pi/3, 1)` is where our tangent line touches.

For the slope, we must calculate the derivative of the function, and then find the exact derivative at `pi/3`

`(df)/dx = secxtanx + 2sinx`

If `x=pi/3...`

`(df)/dx (x=pi/3) = sec (pi/3) tan (pi/3) + 2 sin (pi/3) = 1/(1/2) * ((sqrt3)/2)/(1/2) + 2(sqrt3)/2 = 2*sqrt3 + sqrt3 = 3sqrt3`

Thus, our tangent line is of the form `y=(3sqrt3) x + c_1`

Since `y(pi/3)=1...`, use point slope form...

`y-1 = 3sqrt3 (x-pi/3) = 3sqrt3x - pisqrt3 -> y = 3sqrt3x + 1 - sqrt3pi`

Thus, the tangent line equation is:

`y = (3sqrt3)x + 1 - (sqrt3)pi`