Chapter 7.2, Problem 47E

To determine

To find: the equation of sine function.

Answer to Problem 47E

  $y=\pm 2\sin \left(2\theta -\frac{\pi }{2}\right)$

Explanation of Solution

Given information:

  $\begin{array}{l}\text{Amplitude, }\left|A\right|=2\\ \text{Period, }\left(\raisebox{1ex}{$2\pi $}\!\left/ \!\raisebox{-1ex}{$k$}\right.\right)=180°\\ \text{Phase shift, }\left(\raisebox{1ex}{$-c$}\!\left/ \!\raisebox{-1ex}{$k$}\right.\right)=45°\end{array}$

Calculation:

The equation of sine function is as follows:

  $y=A\sin \left(k\theta +c\right)$    … (1)

Here $A$ is coefficient of the absolute value of amplitude.

The value of A is evaluated as follows:

  $\begin{array}{c}A=\pm \left|A\right|\\ =\pm 2\end{array}$

The value of $k$ is evaluated as follows:

  $\begin{array}{c}\text{Period}=\frac{2\pi }{k}\end{array}$

  $\begin{array}{c}180°=\frac{2\pi }{k}\left[\text{Put the value of period}\right]\end{array}$

  $\begin{array}{c}\pi =\frac{2\pi }{k}\left[\text{Convert degree measure into radian}\right]\end{array}$

  $\begin{array}{c}k=\frac{2\pi }{\pi }\end{array}$

  $\begin{array}{c}k=2\end{array}$

The phase shift is evaluated as follows:

  $\begin{array}{c}\text{Phase shift}=\frac{-c}{k}\end{array}$

  $\begin{array}{c}45°=\frac{-c}{2}\left[\text{Put the value of phase shift and }k\right]\end{array}$

  $\begin{array}{c}\left(45°\right)\left(\frac{\pi }{180°}\right)=\frac{-c}{2}\left[\text{Convert degree measure into radian}\right]\end{array}$

  $\begin{array}{c}\frac{\pi }{4}=\frac{-c}{2}\end{array}$

  $\begin{array}{c}\frac{\pi }{4}\left(2\right)=-c\end{array}$

  $\begin{array}{c}c=-\frac{\pi }{4}\left(2\right)\end{array}$

  $\begin{array}{c}=-\frac{\pi }{2}\end{array}$

Put the values of $A,k$ and $c$ in equation (1) to get the equation of sine function as follows:

  $y=\pm 2\sin \left(2\theta -\frac{\pi }{2}\right)$