Proving the bounds of the cosine sequence without starting with basic trig identities.

Question

If we look at a unit circle we can see that the values of cosine are between -1 and 1. However is there a particular proof for this fact? I have tried using the Euler's identity to arrive at a proof with no luck. I have also tried using the expanded form of cosine : $\cos(x)=\sum_{n=0}^{\infty } \frac{(-1)^n x^{2n}}{2n!}$. Using the expansion I tried to manipulate as much as I could however it seems that I am not finding a proof in this fashion either. I also tried going from the fact that it is a Cauchy sequence. I know that identities can be brought in to prove it, but I should like to find a more fundamental way of proving it. Any ideas?

Answer 1

$\cos^2x+\sin^2x=(\cos x+i\sin x)(\cos x -i\sin x)=(\cos x+i\sin x)(\cos (-x )+ i \sin (-x))$

$=e^{ix}e^{-ix}=e^0=1.$

Since $\cos x$ and $\sin x$ are real numbers, their squares are non-negative.

Therefore, $\cos^2x\le1$, which implies $-1\le \cos x\le1$.

Answer 2

Let us write the power series for the cosine as $c(x)$. Since the radius of convergence is infinite we can differentiate term by term. The derivative is $-s(x)$, where $s(x)$ denotes the series for the sine. And $s'(x)=c(x)$.

From $s(x)c(x)-c(x)s(x)=0$ we get $$(c^2(x)+s^2(x))'=0,$$ so it's constant. Evaluation at zero gives $$c^2(x)+s^2(x)=1.$$ In particular $c^2(x)\leq1$, so $|c(x)|\leq1$.