I noticed something just now. This is probably a stupid question, but I'm going to ask it anyway. Because when I discover that my understanding of a topic is fundamentally flawed, I get nervous. 

![enter image description here](https://i.stack.imgur.com/t2fyC.jpg)

Basically I'm suppose to show that the angle marked in red is $sin \space \alpha = \frac{3}{5}$. Note that this task is on the part of the test without calculator. My first thought was that the whole thing is 90 degrees. And the other to angles I can fine easily. AB is 1 and BE is 0.5. And the length of AE is $\frac{\sqrt 5}{2}$. So I calculate the angle for the bottom triangle. 

$sin \space = \frac{BE}{AE} = \frac{\frac{1}{2}}{\frac{\sqrt 5}{2}} = \frac{1}{\sqrt 5}$ 

I know that sine of 90 is 1, right. Now it all falls apart, the following is wrong. 
The angle on the top side of the red angle is equal to the one just calculated. So I did this. 

$1 - 2 \times \frac{1}{\sqrt 5}$  

And expected to get $\frac{3}{5}$, which I didn't. The following is correct. 

$\arcsin(1) - 2 \times \arcsin(\frac{1}{\sqrt 5}) = 36.86$  

$\arcsin(\frac{3}{5}) = 36.86$ 

Why won't the expression without arcsin give me $\frac{3}{5}$ ? 
Hope this makes sense, I'll be right here pressing F5 and updating if more info is needed. Thank you for input.