Question

How do you find the equation of the tangent line `y=cosx` at `(pi/4, sqrt2,2)`?

Answer 1

`y-sqrt(2)/2 = -sqrt(2)/2*(x-pi/4)`

Explanation:

guessing you meant `(pi/4,sqrt(2)/2)`
find `y'(x): d/dx(cosx)`
`= -sinx` (trig differentiation)

slope at `(pi/4,sqrt(2)/2)`:
`y'(pi/4) = -sin(pi/4)`
`= -sqrt(2)/2`

plug into: `y-sqrt(2)/2 = slope*(x-pi/4)`
`y-sqrt(2)/2 = -sqrt(2)/2*(x-pi/4)`
you can convert to slope-intercept if you want