Chapter 6.4, Problem 12CFU

To determine

the equation of the sine function.

Answer to Problem 12CFU

  $y=\pm 0.8\sin \left(2\theta \right)$

Explanation of Solution

Given information:

Amplitude $\left|A\right|$ = 0.8

Period = $\pi$

Calculation:

The equation of sine function is as follows:

  $y=A\sin k\theta$..........(1)

Here $A$ is an amplitude which is an absolute value of the coefficient and $k=\frac{2\pi }{\text{Period}}$ .

Put the values of $A\text{ and }k$ in equation (1) and solve as follows:

  $\begin{array}{l}y=\pm \left|0.8\right|\sin \left(\left(\frac{2\pi }{\text{Period}}\right)\theta \right)\left[\text{Put }A=\pm \left|A\right|\right]\\ y=\pm 0.8\sin \left(\left(\frac{2\pi }{\pi }\right)\theta \right)\left[\text{Put the value of period}\right]\\ y=\pm 0.8\sin \left(2\theta \right)\end{array}$

Thus the expression of sine function is $y=\pm 0.8\sin \left(2\theta \right)$ .