How do you evaluate #cos(arcsin(-2/3)) #?

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Answer 1 · 2015-06-28T23:12:01

Use #sin^2+cos^2=1# to get:

#cos(arcsin(-2/3)) = sqrt(5)/3#

Let #alpha = arcsin(-2/3)#

Then #sin(alpha) = -2/3#

#cos(alpha) = +-sqrt(1 - sin^2(alpha))#

#= +-sqrt(1-(2/3)^2)#

#= +-sqrt(1-4/9)#

#= +-sqrt(5/9)#

#= +-sqrt(5)/3#

By the definition of #arcsin#, #-pi/2 <= alpha <= pi/2#

so #cos(alpha) >= 0#

So we need to pick the positive square root and find:

# cos(arcsin(-2/3)) = cos(alpha) = sqrt(5)/3#