Trigonometric relationship (Posted on 2008-09-19)

	Submitted by pcbouhid
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Let arcsin[cos(arcsinx)] = A. The angle A is bounded between 0 and pi/2; that is, 0 =< A =< pi/2, inasmuch as 0 =< cos(arcsinx) =< 1 ( -pi/2 =< arcsinx =< pi/2). Further, sinA = cos(arcsinx). Consequentely: arcsinx = +- (pi/2 - A) and x = sin[+-(pi/2 - A)] = +- cosA. Similarly, if arccos[sin(arccosx)] = B, then 0 =< B =< pi/2 (for 0 =< sin(arccosx) =< 1, since 0 =< arccosx =< pi) and cosB = sin(arccosx). Consequentely: arccosx = pi/2 -+ B, and x = cos(pi/2 -+ B) = +- sinB. Since cosA = sinB(=+- x): A + B = arcsin[cos(arcsin x)] + arccos[sin(arccosx)] = pi/2.

	Subject	Author	Date
	from inside out -- spoiler	Charlie	2008-09-19 13:06:13
	My trig is rusty (spoiler?)	Steve Herman	2008-09-19 12:24:20