Arccos(x) + Arcsin(x) = Pi/2
Date: 02/04/2001 at 08:01:13 From: Susan Subject: Prove Prove that for all x between -1 and 1, cos^-1(x) = pi/2 - sin^-1(x) Please help! Thanks.
Date: 02/04/2001 at 10:49:49 From: Doctor Robert Subject: Re: Prove I hope that you've drawn a figure. Use a unit circle and pick a point on the circle in the first quadrant. Draw the hypotenuse from the origin to this point and then drop the vertical leg to the horizontal axis. Let the angle at the origin be a and the other acute angle be b. Now label the vertical leg x, the hypotenuse 1 and the horizontal leg sqrt(1-x^2). From the triangle: sin(a) = x cos(a) = sqrt(1-x^2) sin(b) = sqrt(1-x^2) cos(b) = x Use the angle-addition formula: sin(a + b) = sin(a)cos(b) + cos(b)sin(a) = x^2 + (1-x^2) = 1 If sin(a + b) = 1, then a + b = pi/2. (Actually we already knew this, since a and b were the acute angles of a right triangle). But a = sin^-1(x) and b = cos^-1(x). So... sin^-1(x) + cos^-1(x) = pi/2 cos^-1(x) = pi/2 - sin^-1(x) - Doctor Robert, The Math Forum http://mathforum.org/dr.math/
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