Chapter 6, Problem 71RE

To determine

Determine the amplitude, period and phase shift of each function and graph at least two periods.

Answer to Problem 71RE

Amplitude of $y=\frac{1}{2}\sin \left(\frac{3}{2}x-\pi \right)$ is $\boxed{amplitude=\frac{1}{2}}$

Periodicity of $y=\frac{1}{2}\sin \left(\frac{3}{2}x-\pi \right)$ is $\boxed{periodicity=\frac{4\pi }{3}}$

Phase shift of $y=\frac{1}{2}\sin \left(\frac{3}{2}x-\pi \right)$ is $\boxed{\frac{2\pi }{3}}$

Explanation of Solution

Given information:

  $y=\frac{1}{2}\sin \left(\frac{3}{2}x-\pi \right)$

Formula used:

For $f\left(x\right)=a.\sin \left(bx-c\right)+d$ the amplitude is $\left|a\right|$

Periodicity of $a.\sin \left(bx-c\right)+d=\frac{\text{Periodicity of }\sin x}{\left|b\right|}$

For $f\left(x\right)=A\cdot g\left(Bx-C\right)+D$ , where $g\left(x\right)$ is one the basic trigonometric functions, $\frac{C}{B}$ is phase shift and D is vertical shift.

Now calculating the amplitude of the function by using $f\left(x\right)=a.\sin \left(bx-c\right)+d$ the amplitude is $\left|a\right|$ formula:

  $\begin{array}{l}y=\frac{1}{2}\sin \left(\frac{3}{2}x-\pi \right)\\ Here,a=\frac{1}{2}\\ amplitude=\left|a\right|\\ amplitude=\left|\frac{1}{2}\right|\\ \boxed{amplitude=\frac{1}{2}}\end{array}$

Now calculating periodicity by using $a\cdot \sin \left(bx+c\right)+d=\frac{\text{Periodicity of }\sin x}{\left|b\right|}$ formula:

  $\frac{\text{Periodicity of sinx}}{\left|b\right|}$

Periodicity of $\sin x$ is $2\pi$ and $b=\frac{3}{2}$ , therefore:

  $\begin{array}{l}=\frac{2\pi }{\left|\frac{3}{2}\right|}\\ \boxed{periodicity=\frac{4\pi }{3}}\end{array}$

Now calculating the phase shift of $y=\frac{1}{2}\sin \left(\frac{3}{2}x-\pi \right)$ :

  $\begin{array}{l}f\left(x\right)=\frac{1}{2}\sin \left(\frac{3}{2}x-\pi \right)\\ g\left(Bx-C\right)=\sin \left(\frac{3}{2}x-\pi \right),B=\frac{3}{2},C=\pi ,D=0\end{array}$

Therefore, the phasing shift $\frac{C}{B}$ is $\boxed{\frac{\pi }{\frac{3}{2}}=\frac{2\pi }{3}}$