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Trigonometric relationship (Posted on 2008-09-19) Difficulty: 2 of 5
Find the relationship between arcsin[cos(arcsin x)] and arccos[sin(arccos x)].

  Submitted by pcbouhid    
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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) 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.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Solutionfrom inside out -- spoilerCharlie2008-09-19 13:06:13
Some ThoughtsMy trig is rusty (spoiler?)Steve Herman2008-09-19 12:24:20
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