Chapter 5.3, Problem 44E

Solution

To write: The function $y=5\sin x-12\cos x$ into sinusoid form, $y=a\sin \left(bx+c\right)$ .

The sinusoid form is $y=13\sin \left(x-1.176\right)$ .

Given information:

The function is $y=5\sin x-12\cos x$ .

Formula used:

Use formula of sine of sum.

  $\sin \left(u+v\right)=\sin u\cos v+\cos u\sin v$

Calculation:

Rewrite $y=a\sin \left(bx+c\right)$ as follows.

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

Substitute $5\sin x-12\cos x$ for $y$ in the above equation.

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

Compare the coefficients of the above equation.

  $\begin{array}{c}b=1\\ a\cos c=5\\ a\sin c=-12\end{array}$

Substitute the $5$ for $a\cos c$ and $-12$ for $a\sin c$ to find the coefficient $a$ as follows.

  $\begin{array}{c}{\left(a\cos c\right)}^2+{\left(a\sin c\right)}^2={\left(5\right)}^2+{\left(-12\right)}^2\\ a^2\left({\cos }^{2}c+{\sin }^{2}c\right)=25+144\\ a^2\left(1\right)=169\\ a=\pm \sqrt{169}\end{array}$

Take positive value of $a$ .

  $\begin{array}{c}a=\sqrt{169}\\ =13\end{array}$

Find the coefficient $c$ as follows.

  $\begin{array}{c}13\cos c=5\\ \cos c=\frac{5}{13}\\ c={\cos }^{-1}\left(\frac{5}{13}\right)\\ =1.176\text{rad}\end{array}$

Or,

  $\begin{array}{c}13\sin c=-12\\ \sin c=-\frac{12}{13}\\ c=-{\sin }^{-1}\left(\frac{12}{13}\right)\\ =-1.176\text{rad}\end{array}$

Here, $\cos c>0$ and $\sin c<1$ . So, take $c=-1.176$ .

Substitute $13$ for $a$ , $1$ for $b$ and $-1.176$ for $c$ in equation $y=a\sin \left(bx+c\right)$ .

  $\begin{array}{c}y=13\sin \left(1x+\left(-1.176\right)\right)\\ y=13\sin \left(x-1.176\right)\end{array}$

Therefore, the sinusoid form is $y=13\sin \left(x-1.176\right)$ .