Prove that cosx is continuous

For any real number c, we have lim_x→ccosx = cosc and cosx is defined for all real numbers. Thus, the function cosx is continuous everywhere.

Now, we will prove that cosx is continuous for all values of x by the epsilon-delta method. We will use the following two formulas:

1. $\cos x -\cos y=2\sin \dfrac{x+y}{2} \sin \dfrac{y-x}{2}$
2. $|\sin x| \leq |x|$

Prove cosx is continuous

Let f(x)=cosx and let x=c be an arbitrary real number.

Let ε>0 be any given positive number. We need to find a positive δ such that

|f(x)-f(c)| < ε whenever 0<|x-c|<δ

Choose δ=ε.

Now, for $x \in \{0<|x-c| <\delta=\epsilon\}$, we have that

$|f(x)-f(c)| = |\cos x -\cos c|$

$=2 |\sin \dfrac{x+c}{2} \sin \dfrac{c-x}{2}|$ by the above Formula 1.

$\leq 2 |\sin \dfrac{c-x}{2}|$ as we know that sinθ≤1 for all values of θ.

$\leq 2 |\dfrac{x-c}{2}|$ by the above Formula 2.

= |x-c|

< ε.

This shows that for any given ε>0, there exists a δ>0 such that whenever 0<|x-c|<δ, we have

|f(x)-f(x)| < ε.

Thus, by the epsilon-delta definition of continuity, we conclude that cosx is a continuous function at x=c. As c was an arbitrary real number, cosx is everywhere continuous.

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