Is f(x) Constant for -1 ≤ x ≤ 1?

[snipez90]

Homework Statement

Show that f(x) = arcsin(x) + arccos(x) = constant, for $$-1 \leq x \leq 1$$

Homework Equations

Some theorem involving continuity or inverse functions

The Attempt at a Solution

The theorem is easy for the interval with strict bounds (differentiability on open interval and we get f'(x) = 0, so f constant on that interval), but how do I show this for the end points? The hint is to "show and use continuity" but I'm pretty sure I need to use a theorem about inverse functions. I'm just not sure how to deal with the end points -1 and 1. Any help is appreciated.

 

[Office_Shredder]

arcsin(x) and arccos(x) are continuous on the closed interval [-1,1] so arcsin(x) + arccos(x) is too. Then you can take a sequence of points converging to, say, 1 (you do -1 as a separate case). You use the fact that arcsin(x) + arccos(x) is continuous plus the fact that you know the value of arcsin(x) + arccos(x) for each point in the sequence

 

[Mark44]

Here is a simpler approach that uses only basic trig and the ideas of inverse functions.
Let w = arcsin(y)

Draw a right triangle with angle w at left, and right angle at right, with altitude y and hypotenuse 1. The remaining angle, at the top, is (pi/2 - y).
Then sin(w) = y and cos(pi/2 -w) = y.
So w = arcsin(y) and pi/2 - w = arccos(y), hence arcsin(y) + arccos(y) = pi/2, a constant.

The range that is common to both arcsin and arcsin is [0, pi/2], so the representation of w as an angle in the first quadrant is, I believe, all we need.