Question

What is the equation of the tangent line of `f(x) =arcsin(cosx)` at `x=pi/4`?

Answer 1

`y=-x+pi/2`

Explanation:

Know that: `d/dx[arcsin(u)]=(u')/(sqrt(1-u^2))`

Thus:

`d/dx[arcsin(cosx)]=(d/dx[cosx])/(sqrt(1-cos^2x))`

`=(-sinx)/sqrt(sin^2x)=-sinx/abs(sinx)`

To find the slope of the tangent line when `x=pi/4`, find `f'(pi/4)`.

`f'(pi/4)=-sin(pi/4)/abs(sin(pi/4))=-(sqrt2/2)/(sqrt2/2)=-1`

The `y`-coordinate point that the tangent line will touch is:

`f(pi/4)=arcsin(cos(pi/4))=arcsin(sqrt2/2)=pi/4`

`(pi/4,pi/4)`

Write the equation in point-slope form:

`y-pi/4=-(x-pi/4)`

In point-intercept form:

`y=-x+pi/2`

Although unrelated, the appearance of this graph is very interesting, and shows why the absolute value signs were necessary to include.

The internal Socratic grapher won't graph the function, but I would recommend trying it out for yourself here: https://www.desmos.com/calculator