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Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

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  • If $f$ is a real function on a subset of the real numbers and $c$ be a point in the domain of $f$, then $f$ is continuous at $c$ if $\lim\limits_{\large x\to c} f(x) = f(c)$.
Step 1:
Let us first discuss the continuity of cosine.
$f(x)=\cos x$
At $x=c\in R$
$\lim\limits_{\large x\to c}\cos x=\cos c=f(c)$
$f$ is continuous for all values of $x\in R$
Step 2:
Let us discuss the continuity of cosecant.
$f(x)=cosec \;x$
$f$ is not defined at $x=n\pi$
$\Rightarrow f$ is not continuous at $x=n\pi$
Step 3:
Next let us discuss the continuity of secant.
Let $f(x)=\sec x$
$\sec x$ is undefined at $x=\large\frac{(2n+1)\pi}{2}$$\;\;n\in z$
Also at $x=\large\frac{\pi}{2}$
LHL=$\lim\limits_{\large x\to \Large\frac{\pi}{2}}\sec x$
$\quad=\lim\limits_{\large h\to 0}\sec(\large\frac{\pi}{2}$$-h)$
$\quad=\lim\limits_{\large h\to 0}cosec \;h$
RHL=$\lim\limits_{\large x\to \Large\frac{\pi}{2}}\sec x$
$\quad=\lim\limits_{\large h\to 0}\sec(\large\frac{\pi}{2}$$+h)$
$\quad=-\lim\limits_{\large h\to 0}cosec \;h$
RHL $\neq$ LHL
$f$ is not continuous at $x=\large\frac{\pi}{2}$ or at $x=\large\frac{(2n+1)\pi}{2}$
At $x=c\neq \large\frac{(2n+1)\pi}{2}$
$\lim\limits_{\large x\to c}\sec x=\sec c=f(c)$
Hence $f$ is continuous at $x\in R$ except at $x=\large\frac{(2n+1)\pi}{2}$ where $n\in z$
Step 4:
Finally we discuss the continuity of cotangent .
$f(x)=\cot x$
$f$ is not defined at $x=n\pi$
At $x=\pi$
LHL=$\lim\limits_{\large x\to \pi}\cot x$
$\quad\;\;\;=\lim\limits_{\large h\to 0}\cot (\pi-h)$
$\quad\;\;\;=\lim\limits_{\large h\to 0}(-\cot h)$
RHL=$\lim\limits_{\large x\to \pi}\cot x$
$\quad\;\;\;=\lim\limits_{\large h\to 0}\cot (\pi+h)$
$\quad\;\;\;=\lim\limits_{\large h\to 0}(\cot h)$
Thus this $f(x)$ does not exist at $x=n\pi$
At $x=c\neq n\pi$
$\lim\limits_{\large x\to c}\cot x=\cot c=f(c)$
$f$ is continuous at all points $x\in R$ except $x=n\pi$ where $n\in z$
answered May 29, 2013 by sreemathi.v
edited May 29, 2013 by sreemathi.v