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c) sin[cos^-1(-(sq root 3)/2)]


d) cos^-1[cos(2pi/3)]


e) cos[sin^-1(1/2)]

2007-10-28 04:49:03 · 2 answers · asked by Anonymous in Science & Mathematics Mathematics

2 answers

c) sin[cos^-1(-(sq root 3)/2)] = sin(5pi/6) = 1/2

d) cos^-1[cos(2pi/3)] = 2pi/3

e) cos[sin^-1(1/2)] = cos(pi/6) = (sq root 3)/2
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To the answer below me,
Thank you.

2007-10-28 04:52:08 · answer #1 · answered by sahsjing 7 · 1 0

sin(5pi/6) = 1/2 not -1/2
Everthing else that Sahsjing answered looks okay.

2007-10-28 12:06:28 · answer #2 · answered by Demiurge42 7 · 0 0

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