How do you evaluate $cos (arcsin (- \frac{2}{3}))$ $cos(arcsin(-2/3))$ ?

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Answer 1 · 2015-06-28T23:12:01

Use ${sin}^{2} + {cos}^{2} = 1$ $sin^2+cos^2=1$ to get:

$cos (arcsin (- \frac{2}{3})) = \frac{\sqrt{5}}{3}$ $cos(arcsin(-2/3)) = sqrt(5)/3$

Let $α = arcsin (- \frac{2}{3})$ $alpha = arcsin(-2/3)$

Then $sin (α) = - \frac{2}{3}$ $sin(alpha) = -2/3$

$cos (α) = \pm \sqrt{1 - {sin}^{2} (α)}$ $cos(alpha) = +-sqrt(1 - sin^2(alpha))$

$= \pm \sqrt{1 - {(\frac{2}{3})}^{2}}$ $= +-sqrt(1-(2/3)^2)$

$= \pm \sqrt{1 - \frac{4}{9}}$ $= +-sqrt(1-4/9)$

$= \pm \sqrt{\frac{5}{9}}$ $= +-sqrt(5/9)$

$= \pm \frac{\sqrt{5}}{3}$ $= +-sqrt(5)/3$

By the definition of $arcsin$ $arcsin$ , $- \frac{π}{2} \leq α \leq \frac{π}{2}$ $-pi/2 <= alpha <= pi/2$

so $cos (α) \geq 0$ $cos(alpha) >= 0$

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

$cos (arcsin (- \frac{2}{3})) = cos (α) = \frac{\sqrt{5}}{3}$ $cos(arcsin(-2/3)) = cos(alpha) = sqrt(5)/3$

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