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Date: 08/23/97 at 11:04:12
From: Pierre-Marc
Subject: Arccos

Solve this:

       Arc cos (cos y) + Arc sin (sin x) = y + x

Thank you!

Date: 08/24/97 at 08:18:48
From: Doctor Jerry
Subject: Re: Arccos

Hi Pierre-Marc,

I'm not sure I have the patience to think through the special cases.  
However, I graphed the function 

f(x,y) = arccos(cos(y)) + arcsin(sin(x))-y-x

on [0,10]x[0,10] to gain an understanding of what's happening.  
It shows f(x,y) = 0 only near zero, which more or less fits my gut 
feeling about this equation.


I hope this helps.  Perhaps others will offer a solution.

-Doctor Jerry,  The Math Forum
 Check out our web site!   

Date: 08/25/97 at 11:22:26
From: Doctor Wilkinson
Subject: Re: Arccos

    y = y0 + n pi
    x = x0 + m pi

        0 <= y0 < pi
    -pi/2 <= x0 < pi/2

and m and n are integers.  In particular, if m = n = 0, then

    arccos(cos y) + arcsin(sin x) = y + x

which gives us one set of solutions.

In general, we have the equations

    arccos(cos y) = arccos(cos(y0 + n pi)) = pi - y0 if n is odd
        = y0      if n is even
    arcsin(sin y) = arcsin(sin(x0 + m pi)) = -x0     if m is odd
                                           =  x0     if m is even

Now it is convenient to distinguish four cases.  I'll do two of them 
and let you do the other two:

    if m and n are both even, we get the equation

    y0 + x0 = y0 + n pi + x0 + m pi = y0 + x0 + (m + n) pi

which is true if and only if m + n = 0

If m and n are both odd, we get the equation

    pi - y0 - x0 = x0 + y0 + (m + n) pi
    x0 + y0 + (m + n - 1) pi/2 = 0

-Doctor Wilkinson,  The Math Forum
 Check out our web site!   

Date: 06/08/2002 at 23:54:52
From: Darren Lorent
Subject: Arccos

Is it not true that in the interval [0,pi/2],

   Arc cos(cos(x)) = Arc sin(sin(x)) = x ? 

Then in addition to any other solutions, ALL x and y in the square 

  0 <= x <= pi/2,

  0 <= y <= pi/2 

should be solutions.

Date: 06/10/2002 at 12:42:42
From: Doctor Peterson
Subject: Re: Arccos

Hi, Darren.

Yes, that is part of the solution to which Dr. Wilkinson was 
directing Pierre-Marc. 

As you can see from the graph that Dr. Jerry supplied (of only the 
first quadrant, which unfortunately misses the interesting parts!), 
there is a region near the origin, including your square, where the 
function is zero for all (x,y). Dr. Wilkinson's first case includes 
this region, because his "m+n=0" is true when m=n=0. Since there are 
no restrictions on x and y here other than those assumed, that 

      0 <= y < pi  and 

  -pi/2 <= x < pi/2, 

all x and y in this region are solutions.  His second case represents a line segment, for any choice of m and n.

Have you found the other solutions he points you toward? The graph of 
the whole solution set is very interesting!

-Doctor Peterson,  The Math Forum   
Associated Topics:
College Trigonometry
High School Trigonometry

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