Troubles understanding solution to $\cos(\arcsin(\frac{24}{25}))$ I am having troubles understanding how the answer key to my Pre-Calculus and Trigonometry document got to the answer it did from this task/question:
Find the exact value of the expression $$\cos(\arcsin(\frac{24}{25}))$$
At first, I tried to find the value of $\arcsin(\frac{24}{25})$ and then find the cosine of that value, but it seemed way too hard to do without a calculator. I peeked at the answer key, and I saw this:
Let $y=\arcsin(cos(\frac{24}{25})$ Then, $\sin(y)= \frac {24}{25}$ and $\cos(y)= \frac{7}{25}$

I do not understand how they came to this answer or the route to it. Could someone please guide me in the right direction or show me how the document came to this answer?

The basic definition of "sine" is "opposite side divided by hypotenuse" so they have drawn a right triangle with angle (call it $\theta$) on the left, hypotenuse of length 25 and "opposite side" (the vertical line on the right) of length 24.  So $\sin(\theta)= \frac{24}{25}$ and $\theta= \arcsin\left(\frac{24}{25}\right)$.
By the Pythagorean theorem, then, $x^2+ 24^2= x^2+ 576= 25^2= 625$ where "x" is the length of the "near side", the horizontal line at the bottom of the picture.  So $x^2= 625- 576= 49$.  Since the length of a side must be positive, the length of the "near side" is $\sqrt{49}= 7$ as shown in the picture.  
"cosine" is defined as "near side over hypotenuse" we have $cos\left(sin\left(\frac{24}{25}\right)\right)= cos(\theta)= \frac{7}{25}$.
A more "trigonometric" method of getting that result is to use the fact that $sin^2(\theta)+ cos^2(\theta)= 1$.  Setting $\theta= arcsin\left(\frac{24}{25}\right)$ we have $sin(\theta)= \frac{24}{25}$ and $sin^2(\theta)+ cos^2(\theta)= \left(\frac{24}{25}\right)^2+ cos^2(\theta)= 1$ so $cos^2(\theta)= 1- \frac{576}{625}= \frac{625- 576}{625}= \frac{49}{625}$ and then $cos(\theta)= \sqrt{\frac{49}{625}}= \frac{7}{25}$.
