Why isn't math on the sine of angles the same as math on the angles in degrees? 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.  

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. 
 A: You know that $\alpha+2\beta=90^\circ$. There is no rule that $\sin\alpha+2\sin\beta=1$. In maths it's the single most important thing to stick to given rules and not accidentally "invent" new ones.
For your problem: Maybe writing $\sin\alpha=x/y$ with some auxiliarly lines (e.g. the height on the triangle) and successively derive $x$ and $y$ by a chain of Pythagoras applications.
EDIT: Or better use http://en.wikipedia.org/wiki/Law_of_cosines and derive the sine from cos.
A: A $sin$ of an angle is not the same as an angle, it's a function of the angle. You can add angles:
$$\alpha = \alpha_1+\alpha_2$$
But not the $sin$'s:
$$\sin(\alpha) \neq \sin(\alpha_1)+\sin(\alpha_2)$$
In fact, this is true for most functions, and this property is called "non-additivity".
A: Let $\angle EAB = \alpha_1$. Then $\sin(\alpha+\alpha_1)=2/\sqrt{5}$.
$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$
i.e. $\sin\alpha\cos\alpha_1+\cos\alpha\sin\alpha_1=2/\sqrt{5}$
$\cos\alpha_1=2/\sqrt{5}$ and $\sin\alpha_1=1/\sqrt{5}$ (from figure)
Therefore 
$$2\sin\alpha+\cos\alpha=2\tag1$$ 
(Canceling denominator $\sqrt 5$)
If $\sin\alpha=3/5$ then $\cos\alpha=4/5$, i.e. 
$$2\sin\alpha+\cos\alpha=2\frac{3}{5}+\frac{4}{5}=2\tag2$$
Eq. $(1)$ = Eq. $(2)$
Hence proved...
Also $\sin(\alpha+\beta)\ne\sin\alpha+\sin\beta$.
(Which is called non-linearity, probably the answer to the question is this)
