Sketch the Graph of Φ(x) = 2 sin(2x − π/3) for 0 ≤ x ≤ 7π/5
Question:
Sketch the graph of the following function :
\[ \Phi(x)=2\sin\left(2x-\frac{\pi}{3}\right),\quad 0 \le x \le \frac{7\pi}{5} \]
Solution:
We know that the graph of
\[ y=\sin x \]
is a standard sine curve.
In the function
\[ y=2\sin\left(2x-\frac{\pi}{3}\right) \]
- Amplitude \(=2\)
- Period \(=\dfrac{2\pi}{2}=\pi\)
- Phase shift \(=\dfrac{\pi}{6}\) units to the right
Now calculate some important points:
\[ \begin{aligned} x=0 &\Rightarrow y=2\sin\left(-\frac{\pi}{3}\right) =-\sqrt3\\[8pt] x=\frac{\pi}{6} &\Rightarrow y=2\sin0=0\\[8pt] x=\frac{5\pi}{12} &\Rightarrow y=2\sin\frac{\pi}{2}=2\\[8pt] x=\frac{2\pi}{3} &\Rightarrow y=2\sin\pi=0\\[8pt] x=\frac{11\pi}{12} &\Rightarrow y=2\sin\frac{3\pi}{2}=-2\\[8pt] x=\frac{7\pi}{5} &\Rightarrow y=2\sin\left(\frac{37\pi}{15}\right) \approx 1.99 \end{aligned} \]
Thus the curve passes through the points
\[ \left(0,-\sqrt3\right),\quad \left(\frac{\pi}{6},0\right),\quad \left(\frac{5\pi}{12},2\right),\quad \left(\frac{2\pi}{3},0\right),\quad \left(\frac{11\pi}{12},-2\right) \]
Plot these points and draw a smooth sine curve through them.
Hence, the required graph is shown above.
Graph Features:
- Amplitude = \(2\)
- Period = \(\pi\)
- Phase shift = \(\dfrac{\pi}{6}\) to the right
- Domain = \(0 \le x \le \dfrac{7\pi}{5}\)
- Range = \(-2 \le y \le 2\)