Chapter 4.7, Problem 91E

To determine

Write the 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 graph both forms of the function, to find the graph imply.

Answer to Problem 91E

  $3\cos 2t+3\sin 2t=3\sqrt{2}\sin \left(2t+\frac{\pi }{4}\right)$

Explanation of Solution

Given information:

Given function

  $f\left(t\right)=3\cos 2t+3\sin 2t$ .

Calculation:

We have the expression as $3\cos 2t+3\sin 2t$ ……..(1)

From the given function substituting the values of $A$ and $B$ from expression (1) we get

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

That implies that graph of the following two will be equivalent.

  $3\cos 2t+3\sin 2t=3\sqrt{2}\sin \left(2t+\frac{\pi }{4}\right)$

The following is the graph for $f\left(t\right)=3\cos 2t+3\sin 2t$

  

The following is the graph for $3\sqrt{2}\sin \left(2t+\frac{\pi }{4}\right)$