ArccosDate: 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! http://mathforum.org/dr.math/ Date: 08/25/97 at 11:22:26 From: Doctor Wilkinson Subject: Re: Arccos Let y = y0 + n pi x = x0 + m pi where 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 and 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 or x0 + y0 + (m + n - 1) pi/2 = 0 -Doctor Wilkinson, The Math Forum Check out our web site! http://mathforum.org/dr.math/ 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 http://mathforum.org/dr.math/ |
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