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# The statement is true when the inverse trig functions refer to the principal values, but that requires more careful attention to show than the other answer provides.

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When the inverse trig functions are considered multivalued, we get a more nuanced result, for example

We have to subtract to get #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 #textArctextsin(x) + textArctextcos(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.

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We"ll look at each side of the equivalent

# textArctextcos(x) stackrel?= pi/2 - textArctextsin(x)#

We"ll take the cosine of both sides.

#cos( textArctextcos(x) )= x #

#cos(pi/2 - textArctextsin(x)) = sin(textArctextsin(x)) =x #

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

#cos( textArctextcos(x) )= cos(pi/2 - textArctextsin(x)) #

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

#textArctextcos(x) = pi/2 - textArctextsin(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,#

#textArctextcos(x) = pi/2 - textArctextsin(x) quad#

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

#-pi/2 le textArctextsin(x)

#pi/2 ge - textArctextsin(x) > 0#

#pi ge pi/2 - textArctextsin(x) > pi/2 #

#pi/2

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

# pi/2

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