Chapter 12.7, Problem 43HP

To determine

To find: The Trigonometric function with given conditions.

Answer to Problem 43HP

The function with given conditions is

  $y=3\sin \left(2\theta \right)$

Explanation of Solution

Given information:

Amplitude $=3$

Period $={180}^0$

Concept:

If a function is in the form

  $y=a\sin \left(bx\right)$

Then

Amplitude $=\left|a\right|$

Period $=\frac{2\pi }{\left|b\right|}$

Given

Amplitude $=3$

Period $={180}^0$

  $y=3\sin \left(2\theta \right)$

On comaparing with $y=a\sin \left(bx\right)$

We get

  $\begin{array}{l}a=3\\ b=2\end{array}$

Amplitude of this function $=\left|a\right|=3$

Period $=\frac{2\pi }{2}$

  $\Rightarrow$ Period $=\pi$ or ${180}^0$

Therefore,

The function with given conditions is

  $y=3\sin \left(2\theta \right)$

Graph of this function is