Sketching of \(y=a\sin \left( bx \right)\)
Sketching of \(y=a\sin \left( bx \right)\)
The midline is halfway between the extremes; amplitude is the distance from it to either extreme. The period is one complete cycle.
\(y=\sin x\)
| Function | \(0^\circ\) (\(0\)) | \(90^\circ\) (\(\frac{\pi}{2}\)) | \(180^\circ\) (\(\pi\)) | \(270^\circ\) (\(\frac{3\pi}{2}\)) | \(360^\circ\) (\(2\pi\)) |
|---|---|---|---|---|---|
| \(\sin x\) |
\(y=a\sin(bx)\), \(a,b>0\)
\(\text{Amplitude}=a\qquad T=\dfrac{360^\circ}{b}\text{ or}\hspace{0.5em}\dfrac{2\pi}{b}\)
| Function | Amplitude | Period (degrees) | Period (radians) | Maximum value of \(y\) | Minimum value of \(y\) |
|---|---|---|---|---|---|
| \(y=2\sin 6x\) | |||||
| \(y=5\sin 2x\) | |||||
| \(y=\sin 4x\) | |||||
| \(y=\sin\frac{x}{2}\) |
Mark quarter-period steps and the pattern \(0,a,0,-a,0\). Label the extremes and stop at the stated endpoints.
Example: \(y=3\sin2x\), \(0\le x\le360^\circ\)
Amplitude \(3\); period \(180^\circ\). Quarter-period points: \((0^\circ,0)\), \((45^\circ,3)\), \((90^\circ,0)\), \((135^\circ,-3)\), \((180^\circ,0)\). Repeat for the second cycle.
Video Solution:
Video solution locked
A free account saves your progress; this video needs tuition enrolment.
Solution:
Solution locked
680+ questions here come with free worked solutions. A free account saves your questions and progress; this one needs tuition enrolment.
Create a free account to practise similar questions.
Need help? Join our JC Math tuition classes.
Learn more