Trig proofDate: 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 7). Instead, 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/