A purely analytical proof that for all $-1 \le y \le 1$ there is at least one $x$ such that $\cos(x) = y$ and at least one $z$ such that $\sin(z)=y$ I'm reading Tom Apostol's Calculus, Volume I . . .; there the sine and cosine functions are constructed synthetically from the following axioms:


*

*The sine and cosine function are defined everywhere on the real line.

*$\cos(0)=\sin(\dfrac{\pi}{2})=1$ and  $\cos(\pi)=-1$

*$\cos(x-y)=\cos(x)\cos(y)+\sin(x)\sin(y)$

*if $0 \lt x \lt \dfrac{\pi}{2}$ then $0 \lt \cos(x) \lt \dfrac{\sin(x)}{x} \lt \dfrac{1}{\cos(x)}$


I'm trying to prove that the range of the tangent function is the set of real numbers. This is easy enough to prove geometrically (as well as proving that the proposition from the question title is true) but i've had trouble proving it using only the above axioms and theorems proved only from them; i could do it if i have the proof of the proposition from the question's title, but i'm having trouble proving that as well, specially in regard to when $y$ is the product between $2\pi$ and an irrational number.
 A: Let's repeat the axioms, first, to give them tags for easier reference:
$$\cos(0)=\sin\left(\dfrac{\pi}{2}\right)=1\quad\mbox{and}\quad \cos(\pi)=-1\tag{values}$$
$$\cos(x-y)=\cos x\cos y+\sin x\sin y\tag{subtract}$$
$$0 \lt \cos x \lt \dfrac{\sin x}{x} \lt \dfrac{1}{\cos x}\tag{estimate}$$ for
$0 \lt x \lt \dfrac{\pi}{2}$.
The idea is now to derive the addition theorems and related identities from (subtract) and (values), that's not difficult, but lengthy. Then, we'll derive from (estimate) the inequality $|\sin x|\le|x|$ and use it to prove (uniform!) continuity of $\sin x$ and $\cos x$.
First things first: setting $y=x$ in (subtract) and using (values), we immediately obtain
$$\cos^2x+\sin^2x=1\tag{pyth}$$ and thus the important property
$$|\sin x|\le1\quad\mbox{and}\quad|\cos x|\le1\tag{bounded}$$
Setting $x=0$ in (subtract) and using (values), we get
$$\cos(-y)=\cos y\tag{even}$$
With $x=\frac{\pi}2$ in (subtract) and using (values), we get
$$\cos\left(\frac{\pi}2-y\right)=\sin y\tag{reflect1}$$ and from that, replacing $y$ by $\frac{\pi}2-y$,
$$\sin\left(\frac{\pi}2-y\right)=\cos y\tag{reflect2}$$
Well, it's time for the first addition theorem: replacing in (subtract) $x$ by $\frac{\pi}2-x$ and using (reflect1) and (reflect2), we see
\begin{align}\sin(x+y)&=\cos\left(\frac{\pi}2-x-y\right)\\
&=\cos\left(\frac{\pi}2-x\right)\cos y+\sin\left(\frac{\pi}2-x\right)\sin y\\
&=\sin x\cos y+\cos x\sin y\tag{add1}
\end{align} 
To prove the other one, we first have to prove something we've always known: $\sin x$ is odd. At the same time, we make a first step towards proving that our functions are indeed periodic. But there's still a missing special value: setting $x=\pi$ in (pyth) and using (values), we get
$$\sin\pi=0\tag{values2}$$
Setting $y=\pi$ in (subtract) and using (values) and (values2), we obtain
$$\cos(x-\pi)=-\cos x\tag{halfperiod}$$
Replacing $x$ by $\frac{\pi}2-x$, this becomes $$\cos\left(-\frac{\pi}2-x\right)=-\cos\left(\frac{\pi}2-x\right)$$ But using (even) and (reflect), we see that the LHS is $\sin(-x)$, while the RHS is $-\sin x$, so we get
$$\sin(-x)=-\sin x\tag{odd}$$ Finally, we can replace $y$ by $-y$ in (subtract) to obtain the missing addition theorem, using (even) and (odd):
$$\cos(x+y)=\cos x\cos y-\sin x\sin y\tag{add2}$$
Combining (subtract) and (add2), we have 
$$\cos(x-y)-\cos(x+y)=2\,\sin x\sin y,$$ and replacing $x$ by $(x+y)/2$ and $y$ by $(x-y)/2$, this becomes
$$\cos y-\cos x=2\,\sin\frac{x-y}2\,\sin\frac{x+y}2\tag{diff1}$$ 
In the same way, we can derive $$\sin x-\sin y=2\,\sin\frac{x-y}2\,\cos\frac{x+y}2\tag{diff2}$$
And we still need an important relation for double arguments, and obtain it from (add1) with $y=x$: $$\sin2x=2\,\sin x\,\cos x\tag{double}$$
Up to now, this was sheer algebra, but for analysis, we need inequalites, too. For $0 \lt x \lt \dfrac{\pi}{2}$, we can multiply the right-hand part of (estimate) by the (positive in that region) $\cos x$ to get $$\frac{\sin x\,\cos x}x<1$$ By (double), this means $$\frac{\sin 2x}{2x}<1$$ i.e.
$$\frac{\sin x}{x}<1$$ even in the interval $(0,\pi)$. Due to (bounded), this must be valid for all positive $x$, and since the LHS is an even function, for all $x\neq0$. This means
$$|\sin x|\le|x|\tag{estimate2}$$ for $x\neq0$, but it's obviously valid for $x=0$, too.
Let's use this in (diff1), now: with (estimate2) and (bounded), it becomes
$$|\cos x-\cos y|\le|x-y|\tag{lipschitz1}$$ and analogously from (diff2)
$$|\sin x-\sin y|\le|x-y|\tag{lipschitz2}$$
It follows $\sin x$ and $\cos x$ are Lipschitz continuous (and thus uniformly continuous) on the whole real line.
