Is there a purely algebraic proof to show that $-1\leq\sin x\leq1?$ I have to prove the boundedness of $\sin x$ (strict inequality) ie. 

$-1\leq\sin x\leq1$.

I know a geometric proof using trianglesbut I am not too satisfied with it as it does not prove that $=1$ part properly.(Ax,$\sin x$=$\frac{\text {opposite}}{\text{hypotenuse}}$ and in a real triangle $\text{opposite}\neq\text{hypotenuse}$.)
So,I am looking for a purely algebraic proof which does not use any other trigonometric ration like $\cos x$ etc and assuming it to be less than $1$.
Thanks for any help and response!!
 A: Here is $abc$ triangle with $\sin\theta=\frac{b}{c}.$

We will use only two facts. 


*

*Triangle inequality: $a\leqslant b+c$

*Pythagorean theorem: $a^2+b^2=c^2$


From (1) we have that $b+c\geqslant a\geqslant 0$ and hence $b\geqslant -c.$ Dividing by $c$ we have that $\sin\theta\geqslant -1.$
From (2) we have that $b^2= c^2-a^2\leqslant c^2.$ Hence $b\leqslant c.$ Dividing by $c$ we have that $\sin\theta\leqslant 1.$
A: Let $ f(x) = \sin^2 (x) + \cos^2 (x) $. From the series expansion definition, it is evident that $ f $ is continuous and differentiable. On the other hand, we have
$$ f'(x) = 2\cos(x)\sin(x) - 2\sin(x)\cos(x) = 0 $$
so $ f $ is a constant function. As $ f(0) = 1 $, we conclude that $ f(x) = 1 $ in general.
Now, $ \sin^2 (x) \leq \sin^2 (x) + \cos^2 (x) = 1 $, so the result follows.
A: I guess the most elegant and rigorous way, would be to define $ sin (x)$ as the solution to the boundary value problem
$$ y''+y=0$$
$$y(-\pi)=y(\pi)=0$$
We may need to add more boundary conditions to guarantee uniqueness (I did not check.)
Then all properties of $sin$ and $cosine$, defined as its derivative, should follow from this BVP. And I think we can do a lot. Even prove analyticity.
