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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!!

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    $\begingroup$ What is your definition of $\sin x$? $\endgroup$ – user133281 May 13 '16 at 12:34
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    $\begingroup$ Can you use $\cos^2x+\sin^2x=1$? $\endgroup$ – almagest May 13 '16 at 12:36
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    $\begingroup$ @tatan Why does it assume $\cos x\le 1$? You have $\cos^2x\ge0$, hence $\sin^2x\le 1$, hence $-1\le\sin x\le1$. $\endgroup$ – almagest May 13 '16 at 12:40
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    $\begingroup$ If you defined $ \sin(x) $ as "It is a ratio between the opposite side of a triangle and its hypotenuse", how do we use a series expansion? We could show that the series expansion is valid, but we still would have to resort to some geometry because our very definition is geometrical. $\endgroup$ – Ege Erdil May 13 '16 at 12:41
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    $\begingroup$ So, can we use the limit that $\displaystyle\lim_{x\to 0}\dfrac{\sin x}{x}=1$? $\endgroup$ – user170039 May 13 '16 at 12:41
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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.

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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.

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  • $\begingroup$ Definitely need more conditions for uniqueness; your equation has $y=0$ as a solution, as well as $y=2\sin x$. $\endgroup$ – πr8 May 13 '16 at 14:10
  • $\begingroup$ So, add, $y'=1$ also, at the boundary. But do you agree with me on the possibility of deriving properties of $sin(x)$ from this BVP? $\endgroup$ – Behnam Esmayli May 13 '16 at 16:04
  • $\begingroup$ Yeah - you can show that $y^2+(y')^2$ is a constant by differentiating it, and you use initial conditions to show that the constant is $1$. $\endgroup$ – πr8 May 14 '16 at 2:58
  • $\begingroup$ And then you're done :) sum of two nonnegative numbers is one, so each has to be $\leq 1.$ $\endgroup$ – Behnam Esmayli May 14 '16 at 4:11
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Here is $abc$ triangle with $\sin\theta=\frac{b}{c}.$

enter image description here

We will use only two facts.

  1. Triangle inequality: $a\leqslant b+c$
  2. 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 $\sin\theta\leqslant 1.$

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  • $\begingroup$ This does probably not fall in the OP's category of "purely algebraic proofs". Also, this proof only works for $\;0 \le \theta \le 90^\circ\;$. $\endgroup$ – MarnixKlooster ReinstateMonica May 13 '16 at 21:02
  • $\begingroup$ @MarnixKlooster OP wrote in the second comment that "It is a ratio between the opposite side of a triangle and its hypotenuse". So it is imposible to use only algebra with such definition. However a triangle can be viewed "algebraicly" if we embed it in a cartesian plane with coordinates. In such case we can define $\sin\theta$ for all angles and since $\mathbb{R}^2$ is a Hilbert space (2) works. (1) would also work due to the fact that $\mathbb{R}^2$ is a metric space. $\endgroup$ – Fallen Apart May 13 '16 at 23:11

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