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### Trig proof

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Date: Wed, 30 Nov 1994 17:29:41 -0400 (EDT)
From: NAME NEERA
Subject: Math Question

Hello.  I am a faculty member in math.  I am trying to see how this
Dr.Math stuff works.  Here is my question.  Can you show me how
to prove arcsin(x)+arccos(x)=pi\2?
```

```
Date: Mon, 5 Dec 1994 17:07:41 -0500 (EST)
From: Dr. Sydney
Subject: Re: Math Question

Hello!  Thanks for writing Dr. Math.  We are pretty overwhelmed with
questions from students K-12, but I will try to provide some insight into
your problem.  One way to think about it is this:  Say we are given an x.
Then arcsinx describes a certain angle, call it r.  That means at angle
r,
the coordinates on the unit circle are (z,x), where z=Sqrt[1 - x^2].
arccosx describes a different angle, call it q..  That means at angle q,
the
coordinates on the unit circle are (x,z).  Now draw a right triangle with
sides x, z and hypotenuse 1.  Where can you go from there?

There are many approaches to this problem, and I don't mean to even claim
that the above approach is a rigorous proof... it's just the first thing
I
thought of.  We thank you for your interest in Dr. Math, and please tell
anyone you know in K-12th grade to write us.

--Sydney, "Dr." Math
```

```
Date: 01/12/2001 at 22:21:53
From: Jim Hodge
Subject: Trigonometry

Here is an answer for the person who asked the first question under
trigonometry under the category of "College and Beyond" on your
website. Feel free to post it.

The person asked for a proof of:

arcsin x + arccos x = pi/2

First, for the sake of understanding, restate the expression above as;
"The angle whose sine is x plus the angle whose cosine is that same x
which equals pi/2."

Let arcsin x = a   .............................................[1]
And arccos x = b   .............................................[2]

By definition of arcsin and arccos:

sin a = x   ...............................................[3]

and

cos b = x   ...............................................[4]

Therefore:

sin a = cos b   ...........................................[5]

Now draw a unit circle about the origin of an xy plane. Draw a segment
from the origin to the unit circle, then drop a segment straight to the
x-axis. Label the hypotenuse of the triangle thus formed '1', the angle
between the x-axis and the hypotenuse 'a', and the vertical line
segment 'y'. Label the right angle formed with the x-axis 'pi/2', and
the last remaining angle 'pi/2 - a' (since we know that the sum of the
angles in a triangle = pi). From this diagram, we have:

sin a = y

and

cos (pi/2-a) = y

Therefore

sin a = cos (pi/2 - a) ....................................[6]

Substituting [6] into [5] for sin a, we have:

cos(pi/2 - a) = cos b

from which

pi/2 - a = b

pi/2 = b + a = a + b

So that:  a + b = pi/2   .......................................[7]

Substituting [1] and [2] above into [7] we have:

arcsin x + arccos x = pi/2

To take this one step further:

Look at the unit circle created about the origin of the xy plane and
see that for any point on the circle, the x component is the cosine of
the angle formed and the y component is the sine.  Since we know from
(5) above that the sin a = cos b, we can say we are looking for points
on the unit circle where:

y = x   ...................................................[8]

which is the equation of a line through the origin. The equation of the
unit circle is:

x^2 + y^2 = 1   ...........................................[9]

and we then have two equations and two unknowns.

Substituting [8] into [9]:

x^2 + x^2 = 1
2x^2 = 1
x^2 = 1/2
x = +- 1/sqrt(2)

The angle whose sin and cos are 1/sqrt(2) is pi/4.

pi/4 + pi/4 = pi/2

The angle whose sin and cos are -1/sqrt(2) is 5*pi/4.

5*pi/4 + 5*pi/4 = 10*pi/4
=  8*pi/4 + 2*pi/4
=  2*pi   + 2*pi/4
= 2*pi/4
= pi/2
```

```
Date: 09/15/2001 at 22:30:39
From: Doctor Rick
Subject: Re: Trigonometry

Actually, there is a flaw in the proof: it overlooks the complication
inherent in the multiple-valued nature of arcsin and arccos. One cannot
say that, if cos(a) = cos(c), then a = c, as Jim said between eqs. 6 and

if cos(a) = cos(c)
then a = 2*pi*n + c
or   a = 2*pi*n - c
where n is some integer

For instance, these are all equal, though the arguments are not:

cos(pi/4) = cos(9*pi/4) = cos(7*pi/4) = cos(-pi/4)

To deal with the multi-valued nature of the inverse sine and inverse
cosine, we add a condition to the definitions of arccos and arcsin to
make each of these a function, that is, single-valued. The condition is

arcsin(x) = y if and only if sin(y) = x AND -pi/2 < y <= pi/2
arccos(x) = y if and only if cos(y) = x AND 0 <= y < pi

We call the value y that falls in this range the "principal value" of the
inverse sine or cosine.

Now we must modify the proof to account for the correct definitions.
Following the version in our Archives, we note that by the definitions
above

sin(a) = x, -pi/2 <  a <= pi/2
cos(b) = x,     0 <= b <  pi

Then after substituting in equation 5 to get

cos(pi/2 - a) = cos(b)

we conclude that

b = 2*pi*n + pi/2 - a  OR
b = 2*pi*n - pi/2 + a, for some integer n

How do we narrow down the possibilities? We know that

-pi/2 < a <= pi/2

From this we can conclude that

2*pi*n <= 2*pi*n + pi/2 - a < pi + 2*pi*n, and

2*pi*n - pi < 2*pi*n - pi/2 + a <= 2*pi*n

We know that b is equal to one of the expressions in the middle, so
either

2*pi*n <= b < pi + 2*pi*n  (if b = 2*pi*n + pi/2 - a), or

2*pi*n - pi < b <= 2*pi*n  (if b = 2*pi*n - pi/2 + a)

Since 0 <= b < pi, the first case leads to the conclusions

2*pi*n < pi, so n < 1/2
0 < pi + 2*pi*n, so n > -1/2

therefore n must be 0. Thus, for the first case,

b = pi/2 - a
a + b = pi/2

The second case leads to the conclusions

2*pi*n < 2*pi, so n < 1
0 <= 2*pi*n,   so n >= 0

so again n must be 0; but in this case we find that

-pi < b <= 0

so this case is only possible if b = 0, and a = pi/2. Again,

a + b = pi/2

Thus we have at last closed all the loopholes and proved the theorem that

arcsin(x) + arccos(x) = pi/2, for all x

- Doctor Rick, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Trigonometry

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