Question

How do you find the equation for the lines that are tangent and normal to the curve `y=1+cosx` at the point `(pi/2, 1)`?

Answer 1

Start with the derivative

Explanation:

For `y = 1 + cosx`,
`dy/dx = -sinx`.
At `pi/2`, we have
`dy/dx = -sin(pi/2) = -1`.
This is the slope of the tangent line.

Use the point-slope form of a line to find its equation.
`y - y_1 = m(x - x_1)`

In this case
`y - 1 = -1(x - pi/2)`
Solve for y.
`y = -x + pi/2 + 1`

The normal is perpendicular to the tangent.
So, if `m` is the slope of the tangent line, then `-1/m` is the slope of the normal.

In this case, the normal slope is `1`.

Use the same point, and the point-slope form of a line, to obtain the equation of the normal line.