# Sine And Cosine Are Continuous Functions

Weierstrass' $\epsilon -\delta$ definition of limit and continuity is often a ban at the beginning calculus courses. Its application is usually limited to a linear function which is straightforward. The continuity of polynomials is then obtained by means of the algebraic properties of the limit. $f(t)=\sqrt{t}$ is one other function whose continuity can be proved with relative ease from Weierstrass' definition. Following a 1995 article, I shall apply the $\epsilon -\delta$ definition to proving the continuity of $\sin t$ and $\cos t$ at $t=0$ and then extend the result to other values of $t.$

Function $y=f(t)$ defined in a neighborhood of $t=a$ is continuous at $a$ if $\displaystyle\lim_{t\rightarrow a}f(t)$ exists and is equal to $f(a).$ For $\cos t$ at $t=0,$ this means that we have to establish that $\displaystyle\lim_{t\rightarrow 0}\cos t=1;$ for $\sin t$ we'll need to prove that $\displaystyle\lim_{t\rightarrow 0}\sin t=0.$

We know that $\cos t$ is an *even function* and $\sin t$ odd which allows us to consider only positive $t:$ $t\gt 0.$ For small values, $t$ can be regarded as the angle formed by a ray emanating from the origin with the positive $x\mbox{-axis},$ counted counterclockwise. In this setup, let point $(x,y)$ be located on that ray so that $\displaystyle\cos t=\frac{x}{\sqrt{x^{2}+y^{2}}}$ and $\displaystyle\sin t=\frac{y}{\sqrt{x^{2}+y^{2}}}.$ What we'll take for granted is that, for a fixed $x,$ $y\rightarrow 0$ as $t\rightarrow 0.$ Thus $\displaystyle\lim_{t\rightarrow 0}\cos t=1$ is equivalent to $\displaystyle\lim_{y\rightarrow 0}\frac{x}{\sqrt{x^{2}+y^{2}}}=1$ and $\displaystyle\lim_{t\rightarrow 0}\sin t=0$ is equivalent to $\displaystyle\lim_{y\rightarrow 0}\frac{y}{\sqrt{x^{2}+y^{2}}}=0.$

So, given $\epsilon \gt 0,$ we have to find $\delta\gt 0$ such that $\displaystyle \bigg|\frac{x}{\sqrt{x^{2}+y^{2}}}-1\bigg|\lt\epsilon$ as soon as $0\lt x\lt\delta,$ for cosine. For sine, the condition is $\displaystyle \bigg|\frac{y}{\sqrt{x^{2}+y^{2}}}\bigg|\lt\epsilon.$

The task of find $\delta$ is simplified by the observation that both fractions at hand are less than $1.$ So, for example, for sine, we need to find $\delta$ such that $0\lt y\lt\delta$ will insure

$\displaystyle 0\lt\frac{y}{\sqrt{x^{2}+y^{2}}}\lt\epsilon.$

This is the same as

$\displaystyle 0\lt\frac{y^{2}}{x^{2}+y^{2}}\lt\epsilon^{2},$

or,

$y^{2}(1-\epsilon^{2})\lt \epsilon^{2}x^{2}.$

This, in turn, is equivalent to

$\displaystyle 0\lt y\lt\frac{\epsilon x}{\sqrt{1-\epsilon^{2}}}.$

Thus we can set $\displaystyle\delta =\frac{\epsilon x}{\sqrt{1-\epsilon^{2}}}.$

For cosine the requirement is equivalent to

$\displaystyle 0\lt 1-\frac{x}{\sqrt{x^{2}+y^{2}}}\lt\epsilon$

which again can be solved for $y:$

$\displaystyle y\lt x\frac{\sqrt{1 -(1-\epsilon )^{2}}}{1-\epsilon}.$

The right-hand side of the latter inequality can be taken for the sought $\delta.$ It may be observed that any smaller value of $\delta$ will work as well.

For a generic point $t,$ the continuity of cosine requires $\displaystyle\lim_{\tau\rightarrow 0}\cos (t+\tau )=\cos t;$ for continuity of sine, similarly, we need $\displaystyle\lim_{\tau\rightarrow 0}\sin (t+\tau )=\sin t.$ Let's tackle the case of sine:

$\begin{align}\displaystyle \lim_{\tau\rightarrow 0}\sin (t+\tau )&=\lim_{\tau\rightarrow 0}(\sin t\cdot \cos\tau+\cos t\cdot \sin\tau)\\ &=\sin t\cdot \lim_{\tau\rightarrow 0}\cos\tau+\cos t\cdot \lim_{\tau\rightarrow 0}\sin\tau\\ &=\sin t\cdot 1+\cos t\cdot 0\\ &=\sin t. \end{align}$

### References

- M. J. Pascual (1955), in
*A Century of Calculus*, v. 1, T. M. Apostol et al, MAA, 1992, 117-118

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