# What can you say about the shape of the curve  f(x) = 7 cos( 1/3 x ) + \sqrt{19} sin( 1/3 x )  ?

Jun 28, 2018

The graph of $f$ is a sine wave, also called a sinusoid.

#### Explanation:

A sine wave is described by the function

$f \left(x\right) = A \sin \left(B x + C\right) + D$

where $A$, $B$, $C$ and $D$ are given constants.

We ask; how can we turn our function into this form? Well, notice how our function is of the form

$f \left(x\right) = a \sin u \left(x\right) + b \cos u \left(x\right)$

Where $u \left(x\right)$ is another function in terms of $x$. To make it easier to read, let $u = u \left(x\right)$. Suppose there exists $\omega > 0$ and $\tau$ such that

$a \sin u + b \cos u = \omega \sin \left(u + \tau\right)$

As there is no constant term in the formula for $f \left(x\right)$ and the coefficient of $u$ is $1$, we don't need to add additional constants.

$\omega \left(\sin u + \tau\right) = \omega \sin u \cos \tau + \omega \cos u \sin \tau$

$\textcolor{red}{a} \sin u + \textcolor{b l u e}{b} \cos u = \textcolor{red}{\omega \cos \tau} \sin u + \textcolor{b l u e}{\omega \sin \tau} \cos u$

$\implies \left\{\begin{matrix}\omega \cos \tau = a \\ \omega \sin \tau = b\end{matrix}\right.$

Square both relations and add them to reach the condition:

${\omega}^{2} \left({\sin}^{2} \tau + {\cos}^{2} \tau\right) = {a}^{2} + {b}^{2} \implies \omega = \sqrt{{a}^{2} + {b}^{2}}$

Dividing the second relation by the first yields

$\tan \tau = \frac{b}{a} \implies \tau = \arctan \left(\frac{b}{a}\right)$

Hence

$a \sin u + b \cos u = \sqrt{{a}^{2} + {b}^{2}} \sin \left(u + \arctan b \text{/} a\right)$

$f \left(x\right) = 7 \cos \left(\frac{1}{3} x\right) + \sqrt{19} \sin \left(\frac{1}{3} x\right)$

$= \sqrt{{\left(\sqrt{19}\right)}^{2} + {\left(7\right)}^{2}} \sin \left(\frac{1}{3} x + \arctan 7 \text{/} \sqrt{19}\right)$

$= \sqrt{68} \sin \left(\frac{1}{3} x + \arctan 7 \text{/} \sqrt{19}\right)$

Proving that $f$ defines a sinusoid.