Chapter 12.8, Problem 53PPS

To determine

Find the period of the function and also write the equation using the trigonometric equation.

Answer to Problem 53PPS

The period is ${180}^{\circ }$ and the equation is $y=\sin 2(\theta -45)+3$ .

Explanation of Solution

Given:

It is given in the question that there is graph of a sine function given below:

  

Concept Used:

In this, use the concept of general equation of cosine $y=a\sin b(t-c)+d$ , where a is amplitude, $\frac{2\pi }{\left|b\right|}$ is a period, d is the vertical shift or midline and c is the phase shift and midline is calculate by the average of maximum and minimum value.

Calculation: Let the equation be $y=a\sin b(t-c)+d$ . Since from the graph the minimum is $(90,4)$ and the maximum is $(180,2)$ .

First, calculate the midline by taking average,

  $\begin{array}{l}\frac{4+2}{2}=\frac{6}{2}=3\\ d=3\end{array}$

Now, the amplitude, is the difference between midline and maximum value:

  $\begin{array}{l}4-3=1\\ a=1\end{array}$

Since, the graph takes $90$ units from maximum to minimum which is a half of the period.

Now, it is known that the period is ${180}^{\circ }$ or $\pi$ .

Let’s find b,

  $\begin{array}{l}\pi =\frac{2\pi }{\left|b\right|}\\ b=\frac{2\pi }{\pi }\\ b=2\end{array}$

The phase shift is $45$ units to the right, so $c=45$ .

From the above values put in the equation,

  $y=\sin \theta -5$

Conclusion:

The period is ${180}^{\circ }$ and the equation is $y=\sin 2(\theta -45)+3$ .