Question

How do you find the equation of the tangent and normal line to the curve `y=sinx` at `x=pi/3`?

Answer 1

The tangent is:

`y(x) = sqrt(3)/2+1/2(x-pi/3)`

and the normal is:

`y(x) = sqrt(3)/2-2(x-pi/3)`

Explanation:

Given a (differentiable) curve `y=f(x)` the equation of the tangent line in the point `(x_0, y_0)` where `y_0=f(x_0)` is given by:

`y(x) = f(x_0)+f'(x_0)(x-x_0)`

and consequently the equation of the normal line is:

`y(x) = f(x_0)-1/(f'(x_0))(x-x_0)`

As:

`x_0 = pi/3`
`y_0 =sin(pi/3) = sqrt(3)/2`

`f'(x) = cosx`

`f'(x_0) = cos(pi/3)= 1/2`

We have that the tangent is:

`y(x) = sqrt(3)/2+1/2(x-pi/3)`

and the normal is:

`y(x) = sqrt(3)/2-2(x-pi/3)`