 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Trigonometric relationship (Posted on 2008-09-19) Find the relationship between arcsin[cos(arcsin x)] and arccos[sin(arccos x)].

 Submitted by pcbouhid No Rating Solution: (Hide) 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) andx = 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) andcosB = 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 Please log in:

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