What is the tangent line?

Question

For the function f(x) = cos(2x) at x = pi/6

Answer 1

Slope of the tangent line can be found by taking the derivative of a function at a point. The derivative of f is f'(x) = -2sin(2x) and evaluating at x=π/6 gives f'(π/6) = -2sin(π/3) = -√3.

We now get to use point-slope form to get the equation of the tangent line. Point-slope form is y-y₁=m(x-x₁) where (x₁,y₁) is any point on the line. The point we know is on the line is x=π/6 and y=cos(2(π/6))=1/2 so the equation of the tangent line is y-1/2 = -√3 (x-π/6).

Answer 2

Hi, Adam. Here's how to tackle this problem

We have the function f(x) = cos (2x)

The tangent line touches the curve at x = π/6 => slope of tangent line there = slope of the curve.

The slope of the curve, f, is df/dx = d/dx(cos (2x)) = -2 sin 2x by the Chain Rule

so, df/dx at x = π/6 = - 2 sin (2 . π/6)

= -2 sin π/3

= -2 . √3 / 2

= -√3

Let the equation of the tangent line be: y = mx + b . . . . . . . . . . eqn (1)

The line passes through the point (π/6, cos (2π/6) = (π/6, cosπ/3) = (π/6, 1/2)

So we can plug this point into eqn (1): 1/2 = -√3 x + b [slope m = -√3, from above

=> b = 1/2 + √3

Therefore the tangent line has equation: y = -√3 x + (1/2 + √3)