In this post, we will show that the trigonometric function sinx is continuous for all values of x. Here, we will use the limit method as well as the epsilon-delta definition.
Note that
$\lim\limits_{x \to c} \sin x=\sin c$ and $\sin x$ is defined for all real numbers. Thus, we can say that the function $\sin x$ is continuous everywhere.
Now, we will show that $\sin x$ is continuous at every real number by the epsilon-delta method.
Prove that sinx is Continuous
To prove the sine function is continuous by the epsilon-delta definition, we will use the following two formulas:
- $\sin x -\sin y=2\sin \dfrac{x-y}{2} \cos \dfrac{x+y}{2}$
- $|\sin x| \leq |x|$
Let $f(x)=\sin x$ and $x=c$ be an arbitrary point. 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
Let us choose $\delta=\epsilon$. Now, for $x \in {0<|x-c| <\delta}$, we have that
$|f(x)-f(c)| = |\sin x -\sin c|$
$=2|\sin \dfrac{x-c}{2} \cos \dfrac{x+c}{2}|$ by the above Formula 1.
$\leq 2 |\sin \dfrac{x-c}{2}|$ as we know that $|\cos x| \leq 1$ for all values of x.
$\leq 2|\dfrac{x-c}{2}|$by the above Formula 2.
$=|x-c|$
$<\epsilon$.
Thus, we have shown that for any given $\epsilon>0$, there exists a $\delta>0$ such that whenever $0<|x-c|<\delta$, we have
$|f(x)-f(x)|<\epsilon$.
Hence, by the epsilon-delta definition of Continuity, we conclude that the function $f(x)=\sin x$ is continuous at $x=c$. As the point $c$ was arbitrary, $\sin x$ is everywhere continuous.
Also Read:
Epsilon delta definition of limit
Derivative of 2x by First Principle