Chapter 5, Problem 50RE

Solution

To calculate: Equivalent equation of $y=5\sin 2x-12\cos 2x$ in the form $y=a\sin \left(bx+c\right)$ .

The equivalent equation for $y$ is: $y\approx 13\sin \left(2x-1.18\right)$ .

Given information:

The given equation is $y=5\sin 2x-12\cos 2x$ .

Calculation:

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

  $\begin{array}{c}5\sin 2x+\left(-12\right)\cos 2x=a\left(\sin bx\cos c+\cos bx\sin c\right)\\ =\left(a\cos c\right)\sin bx+\left(a\sin c\right)\cos bx\end{array}$

Comparing the coefficients of both sides of the obtained equation, it is clear that $b=2$ , $a\cos c=5$ , and $a\sin c=-12$ .

Determine the value $a$ of as follows:

  $\begin{array}{c}{\left(a\cos c\right)}^2+{\left(a\sin c\right)}^2=5^2+{\left(-12\right)}^2\\ a^2+{\cos }^{2}c+a^2+{\sin }^{2}c=169\\ a^2\left({\cos }^{2}c+{\sin }^{2}c\right)=169\\ a^2=169\\ a=\pm 13\end{array}$

Consequently, $c={\cos }^{-1}\left(\frac{5}{13}\right)$ or $c={\sin }^{-1}\left(-\frac{12}{13}\right)$ .

Therefore, the equivalent equation for $y$ is:

  $\begin{array}{c}y=5\sin 2x-12\cos 2x\\ =a\sin \left(bx+c\right)\\ =13\sin \left(2x+{\cos }^{-1}\left(\frac{5}{13}\right)\right)\text{or}13\sin \left(2x+{\sin }^{-1}\left(-\frac{12}{13}\right)\right)\\ \approx 13\sin \left(2x-1.18\right)\text{or}13\sin \left(2x-1.18\right)\end{array}$

Draw the graph of $y=5\sin 2x-12\cos 2x$ .

  

Draw the graph of $y\approx 13\sin \left(2x-1.18\right)$ .

  

From the obtained graphs, it can be clearly seen that the graph of the functions $y=5\sin 2x-12\cos 2x$ and $y\approx 13\sin \left(2x-1.18\right)$ is exactly same.