Proving the bounds of cosine

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 Eulers 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.Any ideas?

Answer 1

We can very simply prove this using the identity $\sin^2(x)+\cos^2(x)=1$ derived from Pythagoras and the unit circle. This is equivalent to $\cos^2(x)=1-\sin^2(x)$ and since $\sin(x)\in\mathbb{R},\forall{x}\in\mathbb{R}$, it follows that $\sin^2(x)\in\mathbb{R}^+,\forall{x}\in\mathbb{R}$. Our inequality $\cos^2(x)=1-\sin^2(x)$ then turns into $\cos(x)\leq1$, with equality being achieved when $\sin^2(x)$ is minimum (i.e. when $\sin^2(x)=0$ or $x=0$). Therefore this inequality sets the maximum bound of $\cos(x)$ to be 1.