Chapter 12.8, Problem 50PPS

To determine

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

Answer to Problem 50PPS

The period is ${360}^{\circ }$ and the equation is $y=\sin \theta -5$ .

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 sine $h=a\sin b(t-c)+d$ , 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 $h=a\sin b(t-c)+d$ . Since from the graph the minimum is $(-90,-6)$ and the maximum is $(90,-4)$ .

First, calculate the midline by taking average,

  $\begin{array}{l}\frac{-6-4}{2}=\frac{-10}{2}=-5\\ d=-5\end{array}$

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

  $-4-(-5)=1$

So, $a=1$ .

Since, the graph takes ${180}^{\circ }$ from maximum to minimum which is a half of the period.

Now, it is known that the period is ${360}^{\circ }$ or $2\pi$ .

Let’s find b,

  $\begin{array}{l}2\pi =\frac{2\pi }{\left|b\right|}\\ b=\frac{2\pi }{2\pi }\\ b=1\end{array}$

Since, the function $y=\sin x$ , one of the points on the midline is $(0,0)$ .

In this case, one of the points on the midline is $(0,-5)$ .

This means that the phase shift is $0$ , so $c=0$ .

From the above values put in the equation,

  $y=\sin \theta -5$

Conclusion:

The period is ${360}^{\circ }$ and the equation is $y=\sin \theta -5$ .