Answer
1) Check the continuity of $f(x)=\sin x$
2) Check the continuity of $f(x)=\cos x$
3) Check the continuity of $f(x)$ at $\frac{\pi}{4}$
Work Step by Step
$\sin x$ is continuous on $R$, so it is continuous on $(-\infty,\frac{\pi}{4})$
$\cos x$ is continuous on $R$, so it is continuous on $(\frac{\pi}{4},\infty)$
Therefore, $f(x)$ is continuous on $(-\infty,\frac{\pi}{4})\cup(\frac{\pi}{4},\infty)$
Now we only need to check if $f(x)$ is continuous at $\frac{\pi}{4}$.
By definition, $f(x)$ is continuous at $\frac{\pi}{4}$ if and only if $$\lim\limits_{x\to\frac{\pi}{4}}f(x)=f(\frac{\pi}{4})$$
1) For $x\frac{\pi}{4}$, we have
$\lim\limits_{x\to(\frac{\pi}{4})^+}f(x)=\lim\limits_{x\to(\frac{\pi}{4})^+}\cos x=\cos(\frac{\pi}{4})=\frac{\sqrt2}{2}$
Since $\lim\limits_{x\to(\frac{\pi}{4})^-}f(x)=\lim\limits_{x\to(\frac{\pi}{4})^+}f(x)=\frac{\sqrt2}{2}$
Therefore, $\lim\limits_{x\to\frac{\pi}{4}}f(x)=\frac{\sqrt2}{2}$
Notice that $f(\frac{\pi}{4})=\cos(\frac{\pi}{4})=\frac{\sqrt2}{2}$
Since $\lim\limits_{x\to\frac{\pi}{4}}f(x)=f(\frac{\pi}{4})$, $f(x)$ is continuous at $\frac{\pi}{4}$.
Overall, $f(x)$ is continuous on $(-\infty,\infty)$.
