Why $\sin\left(\frac xy\right)$ is not equal to $\frac{\sin x}{\sin y}$ and why $\sin(x+y)$ is not equal to $\sin x+\sin y$ 
Can someone explain to me why $\sin\left(\frac xy\right)$ is not equal to $\frac{\sin x}{\sin y}$, and as an extension, why this holds true for all trig functions? Also, why is $\sin(x+y)$ is not equal to $\sin x+\sin y$, and why does this holds true for all trig functions? 

I get that this may be because they are functions, but what about the nature of trig functions causes the two to examples above to be not equal?
By the way, can you please keep the explanation very simple please? I am a high school student and may struggle to understand more complex explanations involving proof notation etc. 
 A: Just assume $x=y\neq 0$ then
$$\sin(x/x)=\sin 1\neq \sin x/\sin x=1$$
and 
$$\sin(x+x) =\sin (2x)=2\sin x\cos x\neq \sin x +\sin x=2\sin x$$
A: A second of thinking tells you that no function achieves that !
$$\sqrt{x+y}\ne\sqrt x+\sqrt y,\frac1{x+y}\ne\frac1x+\frac1y,\log(x+y)\ne\log x+\log x,\cdots$$
$$\frac xy+1\ne\frac{x+1}{y+1},\tan\frac xy\ne\frac{\tan x}{\tan y}\cdots$$

There are just two exceptions:
$$a(x+y)=ax+ay$$
and 
$$\left(\frac xy\right)^a=\frac{x^a}{y^a}.$$
So you'd better ask why the linear function is additive and why the power function is multiplicative.

If you want to find all additive functions, i.e. such that
$$f(x+y)=f(x)+f(y),$$ you immediately see that 
$$f(2x)=2f(x)$$ and by induction
$$f(nx)=nf(x).$$
This generalizes to rationals,
$$qf\left(\frac pqx\right)=qpf\left(\frac xq\right)=pf(x)\implies f\left(\frac pqx\right)=\frac pqf(x),$$ and to reals
$$f(rx)=rf(x),$$ but the proof is more technical.
Now, setting $r\to x,x\to1$,
$$f(x)=f(1)\,x=ax.$$

For the multiplicative functions
$$g(xy)=g(x)g(y)$$
consider the function
$$f(x):=\log g(e^x)$$ and observe that it is additive, so that
$$f(x)=\log g(e^x)=ax,$$
$$g(e^x)=e^{ax},$$
$$g(x)=x^a.$$
A: If you are looking at the universal statement:

$\sin(x/y)$ is not equal to $\sin(x)/\sin(y)$, for all $x$ and $y$

This statement is false. 
Let $x=\pi^2/(\pi + 2)$, $y=\pi/2$:
$$x/y=\pi-x$$
$$\sin(x/y)=\sin(x)$$
$$\sin(y)=1$$
$$\therefore \sin(x/y)=\sin(x)/\sin(y)$$
The universal statement:

$\sin(x+y)$ is not equal to $\sin(x)+\sin(y)$, for all $x$ and $y$

is also false, since $\sin(2\pi)+\sin(4\pi)=\sin(2\pi+4\pi)$.
A: Note that when $y=\pi$, $$\sin(x+y)=\sin(x+\pi)=-\sin x$$ but $$\sin x+\sin y=\sin x+\sin \pi=\sin x.$$

Also, when $x=2y$, $$\sin\left(\frac xy\right)=\sin2$$ whch is a constant for all $x,y\neq0$ but $$\frac{\sin x}{\sin y}=\frac{\sin 2y}{\sin y}=\frac{2\sin y\cos y}{\sin y}=2\cos y$$ which varies as $y$ varies.
A: Observe that
$$\sin(x+y)=2\sin\frac{x+y}2\cos\frac{x+y}2$$ and $$\sin(x)+\sin(y)=2\sin\frac{x+y}2\cos\frac{x-y}2.$$
The two expressions are equal when $$x+y=2k\pi$$ or when
$$\cos\frac{x+y}2=\cos\frac{x-y}2,$$
or 
$$x+y=\pm(x-y)+4k\pi$$
which is true with
$$x=2k\pi\text{ or }y=2k\pi.$$
Hence $$\sin(x+y)=\sin x+\sin y$$ on a grid of step $2\pi$ plus all NW-SE diagonals.

The case of the division is much less tractable.
A: I think the other answers don't really address the heart of the question, and are too focused on proving why @Ethan_Chan's equations are false specifically for trigonometric functions. Furthermore, I think @Ethan_Chan's misunderstanding is on a more fundamental level.
$\sin(x)$ is a function, like any other. But why shouldn't every function satisfy $f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}$ or $f(x+y)=f(x)+f(y)$? The reason is that a function can map different inputs ($x$ and $y$) to different outputs ($f(x)$ and $f(y)$). But the relationships between $x$ and $y$ aren't necessarily the same as the relationships between $f(x)$ and $f(y)$.
Let's think about what's happening to $x$ and $y$ in each of your examples. You're performing an operation on them: division ($\frac{x}{y}$) or addition ($x+y$). Let's denote an operation on $x$ and $y$ as $x*y$ (be it addition, subtraction, etc.). This brings us to a general rule for why your examples don't work:
$$\text{in general, }f(x*y)\neq f(x)*f(y)$$
@Yves_Daoust's answer shows us the exceptions to this rule. But in general, it is true.
