Chapter 5, Problem 49RE

Solution

To calculate: Equivalent equation of $y=3\sin 3x+4\cos 3x$ in the form $y=a\sin \left(bx+c\right)$ .

The equivalent equation for $y$ is $y\approx 3\sin \left(3x+0.93\right)$ .

Given information:

The given equation is $y=3\sin 3x+4\cos 3x$ .

Calculation:

Since $a\sin \left(bx+c\right)=a\left(\sin bx\cos c+\cos bx\sin c\right)$ , it is clear that

  $\begin{array}{c}3\sin 3x+4\cos 3x=a\left(\sin bx\cos c+\cos bx\sin c\right)\\ =\left(a\cos c\right)\sin bx+\left(a\sin c\right)\cos bx\end{array}$

Comparing the coefficients of both sides of the obtained equation, it is clear that $b=3$ , $a\cos c=3$ , and $a\sin c=4$ .

Determine the value $a$ of as follows:

  $\begin{array}{c}{\left(a\cos c\right)}^2+{\left(a\sin c\right)}^2=3^2+4^2\\ a^2+{\cos }^{2}c+a^2+{\sin }^{2}c=25\\ a^2\left({\cos }^{2}c+{\sin }^{2}c\right)=25\\ a^2=25\\ a=\pm 5\end{array}$

Consequently, $c={\cos }^{-1}\left(\frac{3}{5}\right)$ or $c={\sin }^{-1}\left(\frac{4}{5}\right)$ .

Therefore, the equivalent equation for $y$ is:

  $\begin{array}{c}y=3\sin 3x+4\cos 3x\\ =a\sin \left(bx+c\right)\\ =5\sin \left(3x+{\cos }^{-1}\left(\frac{3}{5}\right)\right)\text{or}5\sin \left(3x+{\sin }^{-1}\left(\frac{4}{5}\right)\right)\\ \approx 3\sin \left(3x+0.93\right)\text{or}3\sin \left(3x+0.93\right)\end{array}$

Draw the graph of $y=3\sin 3x+4\cos 3x$ (green) and, $y\approx 3\sin \left(3x+0.93\right)$ (blue).

  

From the obtained graph, it can be clearly seen that the graph of the functions $y=3\sin 3x+4\cos 3x$ and $y\approx 3\sin \left(3x+0.93\right)$ is identical.