# 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!!

• What is your definition of $\sin x$? – user133281 May 13 '16 at 12:34
• Can you use $\cos^2x+\sin^2x=1$? – almagest May 13 '16 at 12:36
• @tatan Why does it assume $\cos x\le 1$? You have $\cos^2x\ge0$, hence $\sin^2x\le 1$, hence $-1\le\sin x\le1$. – almagest May 13 '16 at 12:40
• 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. – Ege Erdil May 13 '16 at 12:41
• So, can we use the limit that $\displaystyle\lim_{x\to 0}\dfrac{\sin x}{x}=1$? – user170039 May 13 '16 at 12:41

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.

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.

• Definitely need more conditions for uniqueness; your equation has $y=0$ as a solution, as well as $y=2\sin x$. – πr8 May 13 '16 at 14:10
• 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? – Behnam Esmayli May 13 '16 at 16:04
• 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 $1$. – πr8 May 14 '16 at 2:58
• And then you're done :) sum of two nonnegative numbers is one, so each has to be $\leq 1.$ – Behnam Esmayli May 14 '16 at 4:11

Here is $abc$ triangle with $\sin\theta=\frac{b}{c}.$

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 $\sin\theta\leqslant 1.$

• This does probably not fall in the OP's category of "purely algebraic proofs". Also, this proof only works for $\;0 \le \theta \le 90^\circ\;$. – MarnixKlooster ReinstateMonica May 13 '16 at 21:02
• @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. – Fallen Apart May 13 '16 at 23:11