# Purely algebraic proof of the trigonometric inequalities

While calculating various limits of trigonometric functions, one must resort to the squeeze theorem which is founded on the inequalities $$1 > \frac{\sin x}{x} > \cos x$$ for some "small" $x$. These inequalities are, however, always (to my knowledge) established geometrically by drawing various triangles and circles where one sees that they hold.

Is there a purely algebraic proof of this inequalities, using only the properties of trigonometric functions and not relying on the underlying geometry?

• Of course not! That would beg the question. – user54031 May 15 '15 at 14:30

Both: $$\frac{\sin x}{x}\leq 1,\qquad\frac{\tan x}{x}\geq 1$$ can be seen as convexity inequalities. The first one follows from the fact that $\frac{d}{dx}\sin x = \cos x$ is decreasing on $[0,\pi]$ and the second one follows from the fact that $\frac{d}{dx}\tan x=\frac{1}{\cos^2 x}$ is increasing over the same interval.

• How do you get the fact that ${d\over dx}\sin x=\cos x$ without at some point invoking the very inequality you say it implies? – Barry Cipra May 15 '15 at 15:02
• @BarryCipra: it depends on how $\sin$ and $\cos$ are defined. I usually define them as the imaginary and real part of the complex exponential function. Since $\frac{d}{dx}e^x=e^x$ follows from $e^{x}\cdot e^{y}=e^{x+y}$, we also have $\sin' = \cos$ and $\cos'=-\sin$ since the tangent vector to a unit circle has to be normal to the radius. – Jack D'Aurizio May 15 '15 at 16:12

Are you happy to start from:

$\sin x=x-\frac {x^3}{3!}+\frac{x^5}{5!}-...$

$\frac{\sin x}x=1-\frac {x^2}{3!}+\frac{x^4}{5!}-...$

$\cos x=1-\frac {x}{2!}+\frac{x^4}{4!}-...$

$\cos x-\frac{\sin x}x=-x^2\left(\frac 1{2!}-\frac 1{3!}\right)+x^4\left(\frac 1 {4!}-\frac 1{5!}\right)+...$

• I'd rather not. – user54031 May 15 '15 at 14:25
• I do not think this is an algebraic proof. – xpaul May 15 '15 at 14:31
• This can be considered an algebraic proof. Formal power series are a big part of Algebra, and $\sin x$ and $\cos x$ can be defined formally in this manner. – Joel May 15 '15 at 14:32
• In fact, in Euler's "Analysis of the Infinite", he preferred the algebraic approach to calculus using power series. He felt that this was easier to understand for students of Calculus. His textbook was a standard for Calculus for over one hundred years. – Joel May 15 '15 at 14:34
• The power series of $\sin x$ and $\cos x$ actually predate Calculus. These were discovered by Madhava more than two hundred years before Newton and Leibniz. @xpaul – Joel May 15 '15 at 14:47

Trigonometric identities and other such properties of trig functions are independent of whether angles are measured in radians, degrees, or what have you. But the squeeze theorem for $\sin x\over x$ and other such relations do depend on $x$ being measured in radians, so the underlying geometry is inescapable.

I don't know if it is "algebraic", but you might start from the observation that the function $$\theta\in \mathbb{R}\mapsto e^{i\theta}\in \mathbb{C}$$ is Lipschitz with constant $1$. This can be proved via the fundamental theorem of calculus: $$\left\lvert e^{i\theta+h}-e^{i\theta}\right\rvert=\left\lvert e^{i\theta}\int_0^h ie^{i\, h'}\, dh'\right\rvert\le\lvert h\rvert.$$ So $$\left\lvert\frac{e^{i\theta}-1}{\theta}\right\rvert^2\le 1,$$ and applying Euler's formula one obtains $$\frac{(\cos \theta -1)^2}{\theta^2}+\frac{\sin^2\theta}{\theta^2}\le 1.$$ In particular, $$\left\lvert \frac{\sin \theta}{\theta}\right\rvert\le 1,\qquad\left\lvert\frac{1-\cos\theta}{\theta}\right\rvert\le 1$$ for all $\theta\in \mathbb{R}$.

On some suitable interval around $x$ $$1\geq \cos(x) \geq \cos(x) - x \sin(x)\geq 0.$$ Now integrate on this interval to find for $x\geq 0$ $$x \geq \sin(x) \geq x \cos(x)$$ and the opposite inequalities for $x\leq 0$.