How do you prove #arcsin x + arccos x = pi/2#?

Let
#arcsinx=theta#
then
#x=sintheta=cos(pi/2-theta)#
#=>arccosx=pi/2-theta=pi/2-arcsinx#
#=>arccosx=pi/2-arcsinx#
#=>arcsinx+arccosx=pi/2#

This one is trickier than it looks. The other answer doesn't pay it the proper respect.

A general convention is to use the small letter #arccos(x)# and #arcsin(x)# as multivalued expressions, each respectively indicating all the values whose cosine or sine has a given value #x#.

The meaning of the sum of those is really every possible combination, and those wouldn't always give #pi/2.# They won't even always give one of the coterminal angles #\pi/2 + 2pi k quad# integer #k#, as we'll now show.

Let's see how it works with the multivalued inverse trig functions first. Remember in general #\cos x = cos a# has solutions #x=pm a + 2pi k quad# integer #k#.

# c = arccos x# really means

#x = cos c#

#s = arcsin x# really means

#x = sin s#

#y = s + c#

#x# is playing the role of a real parameter that sweeps from #-1# to #1#. We want to solve for #y#, find all the possible values of #y# which have an #x, s# and #c# that makes these simultaneous equations #x = cos c, x = sin s, y = s+c# true.

#sin s = x = cos c #

#cos(\pi/2 - s) = cos c#

We use our above general solution about the equality of cosines.

# pi/2 - s = \pm c + 2pi k quad # integer #k#

# s \pm c = pi/2 - 2pi k #

So we get the much more nebulous result,

#arcsin x pm arcsin c = pi/2 + 2pi k #

(It's permissible to flip the sign on #k.#)

#----------------#

Let's focus now on the principal values, which I write with capital letters:

Show #text{Arc}text{sin}(x) + text{Arc}text{cos}(x) = pi/2#

The statement is indeed true for the principal values defined in the usual way.

The sum is only defined (until we get pretty deep into complex numbers) for #-1 le x le 1# because the valid sines and cosines are in that range.

We'll look at each side of the equivalent

# text{Arc}text{cos}(x) stackrel{?}{=} pi/2 - text{Arc}text{sin}(x)#

We'll take the cosine of both sides.

#cos( text{Arc}text{cos}(x) )= x #

#cos(pi/2 - text{Arc}text{sin}(x)) = sin(text{Arc}text{sin}(x)) =x #

So without worrying about signs or principal values we're sure

#cos( text{Arc}text{cos}(x) )= cos(pi/2 - text{Arc}text{sin}(x)) #

The tricky part, the part that deserves respect, is the next step:

#text{Arc}text{cos}(x) = pi/2 - text{Arc}text{sin}(x) quad# NOT SURE YET

We have to tread carefully. Let's take the positive and negative #x# separately.

First #0 le x le 1#. That means the principal values of both inverse trig functions are in the first quadrant, between #0# and #pi/2.# Constrained to the first quadrant, equal cosines imply equal angles, so we conclude for #x ge 0,#

#text{Arc}text{cos}(x) = pi/2 - text{Arc}text{sin}(x) quad#

Now #-1 le x < 0.# The principal value of the inverse sign is in the fourth quadrant, and for #x < 0# we usually define the principal value in the range

#-\pi/2 le text{Arc}text{sin}(x) < 0#

#\pi/2 ge - text{Arc}text{sin}(x) > 0#

#pi ge pi/2 - text{Arc}text{sin}(x) > pi/2 #

#pi/2 < pi/2 - text{Arc}text{sin}(x) le pi #

The principal value for the negative inverse cosine is the second quadrant,

# \pi/2 < text{Arc}text{cos}(x) le pi#

So we have two angles in the second quadrant whose cosines are equal, and we can conclude the angles are equal. For #x < 0#,

#text{Arc}text{cos}(x) = pi/2 - text{Arc}text{sin}(x) quad#

So either way,

# text{Arc}text{sin}(x) + text{Arc}text{cos}(x) = pi/2 quad sqrt#