How do you find the exact value of $\cos(\operatorname{arcsin}(\frac{-4}{5}))$? Step by step explanations would be helpful. It's my understanding that the above is equal to $\cos (\theta )$ but I don't know why. Thanks in advance!
 A: Hint1: 
$\cos^{2}\alpha+\sin^{2}\alpha=1$. What does that produce for $\alpha=\arcsin\left(-\frac{4}{5}\right)$?
Hint2:
For any $x\in\left[-1,1\right]$ you have $\arcsin\left(x\right)\in\left[-\frac{1}{2}\pi,\frac{1}{2}\pi\right]$
and for $\alpha\in\left[-\frac{1}{2}\pi,\frac{1}{2}\pi\right]$ you
have $\cos\alpha\geq0$.
A: $\arcsin\left(-\frac{4}{5}\right)$ means the angle $\phi$ for which $\sin(\phi)=-\frac{4}{5}$, so $\cos(\arcsin(-\frac{4}{5}))=\cos(\phi)$
A: Draw yourself a right triangle whose sine is $-\frac{4}{5}$ (which quadrant will it be in)? Then determine the third side of the triangle, and see what the cosine of that angle is.
A: Let $\displaystyle\phi=\arcsin x$
$\displaystyle\implies(i)\sin\phi=x $
and $\displaystyle(ii) -\frac\pi2\le\phi\le\frac\pi2$ based on the definition of principal value of inverse sine
$\displaystyle\implies\cos\phi\ge0\implies \cos\phi=+\sqrt{1-\sin^2\phi}=+\sqrt{1-x^2} $
Put $\displaystyle x=-\frac45$
A: Try considering something like this

A: Consider any right angle triangle.  If we know two sides we can find the third, or in this case if we know the ratio of any two sides we can find the ratio for the third.
From Pythagoras we have
$$\text{hypotenuse}^2 = \text{opposite}^2 + \text{adjacent}^2$$
We want the angle with $\text{arcsin}\left( \frac{-4}{5} \right)$ and since $\sin ( \theta) = \frac{\text{opposite}}{hypotenuse}$ then ignoring signs for now we can find the length of the third side
$$
|\text{opposite}| = 4
$$
$$
|\text{hypotenuse}| = 5
$$
$$
|\text{adjacent}| = \sqrt{|\text{hypotenuse}|^2 - |\text{opposite}|^2} = \sqrt{5^2 - 4^2} = 3
$$
Now by convention since the the principle for arcsin gives angle $\theta$ where:
$- \frac{\pi}{2} \le \theta \le \frac{\pi}{2}$ the cosine will be positive.
$$\cos(\theta) = \frac{\text{adjacent}}{hypotenuse} = \frac{3}{5} = 0.6$$
