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

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dk_ch Share
Mar 12, 2016

Answer:

as shown

Explanation:

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#

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Write your answer here...
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Dean R. Share
May 9, 2018

Answer:

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.

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

#x = sin({3 pi}/4)=cos(pi/4)=1/sqrt{2} quad# but #quad {3pi}/4 + pi/4=pi. #

We have to subtract to get #pi/2#.

Explanation:

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#

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