Chapter 5.3, Problem 45E

Solution

To write: The function $y=\cos 3x+2\sin 3x$ as a sinusoid in the form $y=a\sin \left(bx+c\right)$ .

The sinusoid form is $y=2.236\sin \left(3x+0.4636\right)$ .

Given information:

The function is $y=\cos 3x+2\sin 3x$ .

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 $\cos 3x+2\sin 3x$ for $y$ in the above equation.

  $\begin{array}{c}\cos 3x+2\sin 3x=a\left(\sin bx\cos c+\cos bx\sin c\right)\\ =\left(a\cos c\right)\sin bx+\left(a\sin c\right)\cos bx\end{array}$

Compare the coefficients of the above equation.

  $\begin{array}{c}b=3\\ a\cos c=2\\ a\sin c=1\end{array}$

Substitute $2$ for $a\cos c$ and $1$ for $a\sin c$ to find the coefficient $a$ as follows.

  $\begin{array}{c}{\left(a\cos c\right)}^2+{\left(a\sin c\right)}^2={\left(2\right)}^2+{\left(1\right)}^2\\ a^2\left({\cos }^{2}c+{\sin }^{2}c\right)=4+1\\ a^2\left(1\right)=5\\ a=\pm \sqrt{5}\end{array}$

Take positive value of $a$ .

  $\begin{array}{c}a=\sqrt{5}\\ =2.236\end{array}$

Find the coefficient $c$ as follows.

  $\begin{array}{c}2.236\cos c=2\\ \cos c=\frac{2}{2.236}\\ c={\cos }^{-1}\left(\frac{2}{2.236}\right)\\ =0.4636\text{rad}\end{array}$

Or,

  $\begin{array}{c}2.236\sin c=1\\ \sin c=\frac{1}{2.236}\\ c={\sin }^{-1}\left(\frac{1}{2.236}\right)\\ =0.4636\text{rad}\end{array}$

Substitute $2.236$ for $a$ , $3$ for $b$ and $0.4636$ for $c$ in equation $y=a\sin \left(bx+c\right)$ .

  $y=2.236\sin \left(3x+0.4636\right)$

Therefore, the sinusoid form is $y=2.236\sin \left(3x+0.4636\right)$ .