# Showing that cosine is continous (not uniformly continuous)

I know that there exist many proofs that show $cos(x)$ is uniformly continous that rely on Lipschitz continuity and/or the mean value theorem. However, how can I show that $cos(x)$ is continuous in the first place? The mean value theorem requires that the function be differentiable (hence continuous) in the first place so I can't really use it without proving continuity.

• Use the definition of continuity. – Weijun Zhou Jan 29 '18 at 0:32
• How do you define $\cos$? If you use the Taylor series, then continuity and differentiability follow since the radius of convergence in $\infty$. – copper.hat Jan 29 '18 at 0:35

Use the unit circle. Suppose $0< x < \pi/2$. Then if you progress clockwise on the unit circle distance $x,$ you arrive at the point $(\cos(x), \sin(x))$.

The shortest distance to the origin from this point is $\sin(x)$. That is shorter than the distance traveling along the unit circle, which is $x$. We therefore know $$\sin(x) < x, \qquad 0 < x <\pi/2.$$

You should be able to use the properties of sine and cosine to get to your general result.

• Indeed, using this approach you have $\|(\cos x, \sin x) - (\cos y, \sin y)\| \le |x-y|$ directly. – copper.hat Jan 29 '18 at 0:46

$\begin{array}\\ \cos x-\cos y &=-2\sin((x+y)/2)\sin((x-y)/2)\\ \text{so}\\ |\cos x-\cos y| &=2|\sin((x+y)/2)\sin((x-y)/2)|\\ &\le 2|\sin((x-y)/2)| \qquad\text{since } |\sin| \le 1\\ &\le |x-y| \qquad\text{since }|\sin(z)| \le |z|\\ \end{array}$