Question

How do you find an equation of the tangent line to the curve at the given point `y = cos(2x)` and `x=pi/4`?

Answer 1

`y=-2x+pi/2`

Explanation:

The gradient of tangent is given by the derivative.

So `y=cos(2x)`
`:. dy/dx=-2sin(2x)`

When `x=pi/4 => y=cos(2*pi/4) =cos (pi/2) = 0`
and `x=pi/4 => dy/dx=-2sin (pi/2) -2`

So the tangent at `x=pi/2` has gradient `m=-2` and passes through the point `(pi/4,0)`

Hence, using `y-y_1=m(x-x_1)` we have the required equation is;

`y-0=-2(x-pi/4)`
`y=-2x+pi/2`