Is the correct equation for 62: $y= \sin(\frac{1}{4}x)$ ?

  • $\begingroup$ It looks more like (4/3)sin((1/2)x) but I dont know $\endgroup$– KKZiomekFeb 9, 2017 at 23:45
  • 1$\begingroup$ You have accounted for the period but not the amplitude of the graph. The function is actually $f(x) = 4\sin(\frac{x}{4})$. $\endgroup$ Feb 9, 2017 at 23:56
  • 1$\begingroup$ No, the amplitude of #62 is $A=4$ whereas you have it at $A=1$. The $\frac{x}{4}$ is correct since the period is $P=8\pi$ and $\frac{2\pi}{P}=\frac{1}{4}$. $\endgroup$ Feb 9, 2017 at 23:58
  • $\begingroup$ @KKZiomek: how did you come to this unexpected observation ? $\endgroup$– user65203Feb 10, 2017 at 8:14
  • $\begingroup$ @YvesDaoust I was sleepy, I even don’t remember writing that. o.O Forgive me. $\endgroup$– KKZiomekFeb 10, 2017 at 23:39

4 Answers

A general sinusoidal function incorporating the amplitude $A$, period $P$, phase shift $h$ and vertical shift $k$ takes the form

\begin{equation} f(x) =A\sin\left(\frac{2\pi}{P}(x-h)\right)+k \end{equation}

To determine these values from the graph locate the $y$ value of a “peak”, $y_{peak}$ and the $y$ value of a “valley”, $y_{valley}$ then

\begin{eqnarray} A&=&\frac{1}{2}\left(y_{peak}-y_{valley}\right)\\ k&=&\frac{1}{2}\left(y_{peak}+y_{valley}\right) \end{eqnarray}

The period $P$ is the distance between two adjacent peaks (or two adjacent valleys).

One way to find the phase shift $h$ is to locate the $x$-coordinate $x_L$ of the “valley” nearest the $y$-axis and the $x$-coordinate $x_R$ of the first “peak” to the right of that valley. Then

$$ h=\frac{1}{2}\left(x_R+x_L\right)$$

In Problem (62) we see that

  1. $y_{peak}=4$
  2. $y_{valley}=-4$
  3. $x_L=-2\pi$
  4. $x_R=2\pi$
  5. Peak to peak distance is $P=8\pi$

Therefore

  1. $A=\frac{1}{2}\left(4-(-4))\right)=4$
  2. $k=\frac{1}{2}\left(4+(-4)\right)=0$
  3. $h=\frac{1}{2}\left(-2\pi+2\pi\right)=0$

Recall: The sine wave or sinusoid in its most basic form is: y(t) = A *sin(ωt+θ)

Where:

A = amplitude

w = angular frequency = 2$\pi$f

f = 1/T, where T is period in seconds

θ = phase

t = time (s)

Your function will be of the form: $$y=A\sin(\omega x)$$

The amplitude $A$ is half of the function’s range, it goes from $y=-4$ to $y=4$, therefore the amplitude is: $$A=\frac{4-(-4)}{2}=\frac{8}{2}=4$$

You can obtain the regular period of the function as $T=8\pi$. Therefore, the value of $\omega$ (Angular frequency) should be: $$\omega=\frac{2\pi}{T}=\frac{2\pi}{8\pi}=\frac{1}{4}$$ We can confirm this by plotting it on Desmos Graphing Calculator:

The ordinates go from $-4$ to $4$, so the function must be a sine (or a cosine) times $4$.

It is a sine because it passes through $(0,0)$.

As the roots are on every multiples of $4\pi$, the argument must be $x/4$ (because $4\pi/4=\pi$).

Hence

$$4\sin\left(\frac x2\right).$$

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