Sketch the Graph of f(x) = cot 2x
Question:
Sketch the graph of the following function :
\[ f(x)=\cot2x \]
Solution:
We know that
\[ \cot\theta=\frac{\cos\theta}{\sin\theta} \]
Therefore
\[ f(x)=\cot2x \]
The graph of cotangent decreases from \(+\infty\) to \(-\infty\) in each interval between two consecutive asymptotes.
Whenever
\[ \sin2x=0 \]
the function becomes undefined.
Thus vertical asymptotes occur at
\[ 2x=n\pi \Rightarrow x=\frac{n\pi}{2} \]
Important properties:
- Period \(=\dfrac{\pi}{2}\)
- Vertical asymptotes at \(x=\dfrac{n\pi}{2}\)
- Graph decreases continuously in each interval
Now calculate some important points:
\[ \begin{aligned} x=\frac{\pi}{8} &\Rightarrow y=\cot\frac{\pi}{4}=1\\[8pt] x=\frac{\pi}{4} &\Rightarrow y=\cot\frac{\pi}{2}=0\\[8pt] x=\frac{3\pi}{8} &\Rightarrow y=\cot\frac{3\pi}{4}=-1 \end{aligned} \]
Thus the graph passes through the points
\[ \left(\frac{\pi}{8},1\right),\quad \left(\frac{\pi}{4},0\right),\quad \left(\frac{3\pi}{8},-1\right) \]
The pattern repeats after every interval
\[ \frac{\pi}{2} \]
Plot these points and draw smooth cotangent curves approaching the vertical asymptotes.
Hence, the required graph is shown above.
Graph Features:
- Period \(=\dfrac{\pi}{2}\)
- Vertical asymptotes at \(x=\dfrac{n\pi}{2}\)
- The graph decreases in each interval
- Zeros occur at \(x=\dfrac{\pi}{4},\dfrac{3\pi}{4},\dots\)