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:

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

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