How do you find the exact value of #cos(arcsin(1/3))#?

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Answer 1 · 2016-08-28T06:57:03

#(2sqrt2)/3#

Let #a = arc sin (1/3) in Q1#,

for the principal value, wherein cosine is positive.

The given expression is

#cos a = sqrt (1-sin^2a)= sqrt(1-1/9)=(2sqrt2)/3#.

If Q2 value of #a = 180^o - Q1 a# is permitted, the answer will be

#+-(2sqrt2)//3#,

using

#sin (180^o-a)=sin a# but #cos(180^o-a)=- cos a#.