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.
Best Answer
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
$$f(x)=\log g(e^x)=ax,$$
$$g(e^x)=e^{ax},$$
$$g(x)=x^a.$$