Sketch the Graphs of u(x) = sin²x and v(x) = |sin x| for 0 ≤ x ≤ 2π
Question:
Sketch the graphs of the following functions :
\[ u(x)=\sin^2x,\quad 0 \le x \le 2\pi \]
\[ v(x)=|\sin x|,\quad 0 \le x \le 2\pi \]
Graph of \(u(x)=\sin^2x\)
Solution:
Since
\[ u(x)=\sin^2x \]
therefore all values of the function are non-negative.
Now calculate some important points:
\[ \begin{aligned} x=0 &\Rightarrow y=0\\[6pt] x=\frac{\pi}{2} &\Rightarrow y=1\\[6pt] x=\pi &\Rightarrow y=0\\[6pt] x=\frac{3\pi}{2} &\Rightarrow y=1\\[6pt] x=2\pi &\Rightarrow y=0 \end{aligned} \]
Thus the graph passes through the points
\[ (0,0),\quad \left(\frac{\pi}{2},1\right),\quad (\pi,0),\quad \left(\frac{3\pi}{2},1\right),\quad (2\pi,0) \]
Graph Features:
- Range = \(0 \le y \le 1\)
- Period = \(\pi\)
- The graph always lies above the x-axis
Graph of \(v(x)=|\sin x|\)
Solution:
We know that
\[ v(x)=|\sin x| \]
means the negative part of the sine curve is reflected above the x-axis.
Now calculate some important points:
\[ \begin{aligned} x=0 &\Rightarrow y=0\\[6pt] x=\frac{\pi}{2} &\Rightarrow y=1\\[6pt] x=\pi &\Rightarrow y=0\\[6pt] x=\frac{3\pi}{2} &\Rightarrow y=1\\[6pt] x=2\pi &\Rightarrow y=0 \end{aligned} \]
Thus the graph passes through the points
\[ (0,0),\quad \left(\frac{\pi}{2},1\right),\quad (\pi,0),\quad \left(\frac{3\pi}{2},1\right),\quad (2\pi,0) \]
Graph Features:
- Range = \(0 \le y \le 1\)
- Period = \(2\pi\)
- Negative portion of sine curve is reflected above x-axis