Trigonometric arc functions So, I hope this title is appropriate. I have a test tomorrow... Our teacher kind of blind sided us with this information and I was already drooling by the end of class. Thus, irresponsibly I never bothered to ask for clarification.
So, now that the sorry truth is on the table. Here's one of the problems I'm currently working on and I'm utterly stuck.
$$\cos\left(\arcsin\left(\frac{\sqrt{2}}{2}\right)\right)$$
What do I do?
Thank you.
 A: In this I will show you a general method for solving problems of the type $\operatorname{trig}(\operatorname{arctrig}(a/b))$, where $\operatorname{trig}$ is some trigonometric function ($\cos$, $\sin$, etc), $\operatorname{arctrig}$ is some inverse trig function ($\operatorname{arcsin}$, etc), and $a/b$ is some ratio (if you're given an integer $x$, just use $x/1$).  It involves drawing a right triangle and remembering how the trig functions relate to the ratios of the lengths of the sides of the triangle ($\sin$ is opposite over hypotenuse, etc).
The Method (In the particular case of $\sin(\operatorname{arccos}(a/b))$):
If $\operatorname{arccos}(a/b) = \theta$, then $\cos(\theta) = a/b$.
Using that, you can draw a right triangle and label one of the non-$90^\circ$ angles $\theta$.  From there you can conclude that the adjacent side has length $a$ and the hypotenuse has length $b$.

Using the Pythagorean theorem, you can then see that the opposite side must have length $\sqrt{b^2-a^2}$.
Thus the sine of this angle must be $\dfrac {\sqrt{b^2-a^2}}{b}$.
In your case $a=\sqrt{2}$ and $b=2$, therefore $\sin(\theta) = \dfrac {\sqrt{2^2-\sqrt{2}^2}}{2} = \dfrac {\sqrt{2}}{2}$
A: $\theta = \arcsin(\frac{1}{\sqrt{2}}) = \dfrac{\pi}{4} \Rightarrow \cos \theta = \dfrac{1}{\sqrt{2}}$.
A: When you say that $\theta = \arcsin\,x$, what you are saying is which angle $\theta$ satisfy the equation $x = \sin\,\theta$.
In the case of your exercise, to solve $\arcsin\,\left(\dfrac{\sqrt{2}}{2}\right)$, you have to find the angle which sine is equal to $\dfrac{\sqrt{2}}{2}$. The angles you find are between $-\dfrac{\pi}{2}$ and $\dfrac{\pi}{2}$.
So, $\cos\left(\arcsin\,\left(\dfrac{\sqrt{2}}{2}\right)\right) = \cos\,\left(\dfrac{\pi}{4}\right) = \dfrac{\sqrt{2}}{2}$.
