-7.8.98 amplitude period PS VS graph. of cos eq

In summary, we discussed finding the amplitude, period, phase shift, and vertical shift of a cosine function. We also explored how to graph a cosine function and determined the values of A, B, omega, and phi for a given equation. We also learned that the period can be calculated by dividing 2pi by the value of omega minus phi, and that the value of omega represents the frequency of the function. The phase shift, represented by phi, determines the horizontal shift of the graph.
  • #1
karush
Gold Member
MHB
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Find amplitude, period, PS, VS. then graph.

$[DESMOS]{"version":7,"graph":{"viewport":{"xmin":-10,"ymin":-11.610693119544644,"xmax":10,"ymax":11.610693119544644}},"randomSeed":"996fd79a7f16736ddbff1ce2310a2f50","expressions":{"list":[{"type":"expression","id":"1","color":"#c74440","latex":"y=-3\\cos\\left(\\frac{k \\pi}{2}\\right)+2"}]}}[/DESMOS]$

ok I think these are the plug ins we use
$Y_{cos}=A\cos\left[\omega\left(x-\dfrac{x \phi}{\omega} \right)\right]+B
\implies A\cos\left(\omega x-\phi\right)+B
\implies T=\dfrac{2\pi}{\omega}
\implies PS=\dfrac{\phi}{\omega}$

ok I wanted to do the graph in tikx but was just looking for an pre done one as an example to fit this eq
 
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  • #2
$y = A\cos[(\omega-\phi)x] + B$

$T = \dfrac{2\pi}{\omega-\phi}$

$PS = 0$
 
  • #3
skeeter said:
$y = A\cos[(\omega-\phi)x] + B$

$T = \dfrac{2\pi}{\omega-\phi}$

$PS = 0$
$y=-3\cos\left(\dfrac{x\pi}{2}\right)+2$
so from observation A=|-3|=3 and B=2
$T = \dfrac{2\pi}{\omega-\phi}$
ok I am ? what is $\omega -\phi$

W|A says period is 4

i started a tikz no sure how to transform it ...
$\begin{tikzpicture}[xscale=.5,yscale=.5]
[help lines/.style={black!50,very thin}] \draw[->,thin] (-6,0)--(6,0) node[above] {$x$};
\draw[->,thin] (0,-1)--(0,4) node[above] {$f(x)=sin\ x$};
\node [below] at (-2*3.1416,0) {-2$\pi$};
\node [below] at (-1*3.1416,0) {-$\pi$};
\node [below] at (1*3.1416,0) {$\pi$};
\node [below] at (2*3.1416,0) {2$\pi$};
\draw[very thick,color=red] plot [domain={-360/90}:{360/90},smooth] (\x,{sin(90*\x)});
\end{tikzpicture}$
 
Last edited:
  • #4
karush said:
$y=-3\cos\left(\frac{x\pi}{2}\right)+2$
so from observation A=|-3|=3 and B=2
$T = \dfrac{2\pi}{\omega-\phi}$
ok I am ? what is $\omega -\phi$

W|A says period is 4

$B = (\omega - \phi) = \dfrac{\pi}{2} \implies T = 4$

note $\phi = 0$ for $y=-3\cos\left(\frac{\pi}{2} \cdot x \right)+2$
 
  • #5
skeeter said:
$B = (\omega - \phi) = \dfrac{\pi}{2} \implies T = 4$

note $\phi = 0$ for $y=-3\cos\left(\frac{\pi}{2} \cdot x \right)+2$

so then $\omega=\dfrac{\pi}{2}$
 
  • #6
karush said:
so then $\omega=\dfrac{\pi}{2}$

yes, and $\phi = 0$
 

FAQ: -7.8.98 amplitude period PS VS graph. of cos eq

1. What does the "-7.8.98 amplitude period PS VS graph" represent?

The graph represents a cosine equation, where the amplitude is -7.8, the period is 98, and the phase shift is -7.

2. How is the amplitude of a cosine wave determined?

The amplitude of a cosine wave is the distance from the center line to the peak or trough of the wave. In this graph, the amplitude is -7.8, which means the wave has a peak of 7.8 units above the center line and a trough of 7.8 units below the center line.

3. What does the period of a cosine wave represent?

The period of a cosine wave is the length of one complete cycle of the wave. In this graph, the period is 98, meaning it takes 98 units of time for the wave to complete one full cycle.

4. How does the phase shift affect a cosine wave?

The phase shift of a cosine wave determines the horizontal translation of the wave. In this graph, the phase shift is -7, meaning the wave is shifted 7 units to the right on the x-axis.

5. How can this graph be used in real-world applications?

The cosine equation and its corresponding graph are used in various fields such as physics, engineering, and signal processing to model and analyze periodic phenomena. It can be used to study the behavior of sound waves, electromagnetic waves, and many other natural phenomena.

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