Sketch the Graph of h(x) = 2 sin 3x for 0 ≤ x ≤ 2π/3
Question:
Sketch the graph of the following function :
\[ h(x)=2\sin 3x,\quad 0 \le x \le \frac{2\pi}{3} \]
Solution:
We know that the graph of
\[ y=\sin x \]
is a standard sine curve.
In the function
\[ y=2\sin 3x \]
- Amplitude \(=2\)
- Period \(=\dfrac{2\pi}{3}\)
Thus one complete sine wave is obtained in the interval
\[ 0 \le x \le \frac{2\pi}{3} \]
Now calculate some important points:
\[ \begin{aligned} x=0 &\Rightarrow y=2\sin0=0\\[6pt] x=\frac{\pi}{6} &\Rightarrow y=2\sin\frac{\pi}{2}=2\\[6pt] x=\frac{\pi}{3} &\Rightarrow y=2\sin\pi=0\\[6pt] x=\frac{\pi}{2} &\Rightarrow y=2\sin\frac{3\pi}{2}=-2\\[6pt] x=\frac{2\pi}{3} &\Rightarrow y=2\sin2\pi=0 \end{aligned} \]
Thus the curve passes through the points
\[ (0,0),\quad \left(\frac{\pi}{6},2\right),\quad \left(\frac{\pi}{3},0\right),\quad \left(\frac{\pi}{2},-2\right),\quad \left(\frac{2\pi}{3},0\right) \]
Plot these points and draw a smooth sine curve through them.
Hence, the required graph is shown above.
Graph Features:
- Amplitude = \(2\)
- Period = \(\dfrac{2\pi}{3}\)
- Domain = \(0 \le x \le \dfrac{2\pi}{3}\)
- Range = \(-2 \le y \le 2\)