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| Trigonometric relationship (Posted on 2008-09-19) |
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Find the relationship between arcsin[cos(arcsin x)] and arccos[sin(arccos x)].
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Submitted by pcbouhid
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Solution:
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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.
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