Chapter 4.7, Problem 91E

To determine

To write: The given function in terms of the sine function by using the identity $A\cos \omega t+B\sin \omega t=\sqrt{A^2+B^2}\sin \left(\omega t+\arctan \frac{A}{B}\right)$ and then to graph both forms of function by using a graphing utility and to interpret it.

Answer to Problem 91E

  $f(t)=3\sqrt{2}\sin \left(2t+\frac{\pi }{4}\right)$ , both forms of the function are equivalent.

Explanation of Solution

Given Information: Function is $f(t)=3\cos 2t+3\sin 2t$ .

Calculation:

Comparing the given function with $f(t)=A\cos \omega t+B\sin \omega t$ ,

  $A=3,B=3,\omega =2$ .

Putting these values in given identity,

  $\begin{array}{l}f(t)=3\cos 2t+3\sin 2t\\ =\sqrt{3^2+3^2}\sin \left(2t+\arctan \frac{3}{3}\right)\\ =3\sqrt{2}\sin (2t+\arctan 1)\\ =3\sqrt{2}\sin \left(2t+\frac{\pi }{4}\right)\end{array}$

Thus, $f(t)=3\sqrt{2}\sin \left(2t+\frac{\pi }{4}\right)$ .

Graph:

Graph of both the forms of function is:

:

Interpretation:

As graphs of both forms of function is same, so both forms of the function are equivalent.