MHB -7.8.98 amplitude period PS VS graph. of cos eq

[karush]

Find amplitude, period, PS, VS. then graph.

Spoiler: desos 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

 

[skeeter]

$y = A\cos[(\omega-\phi)x] + B$

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

$PS = 0$

 

[karush]

$y = A\cos[(\omega-\phi)x] + B$

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

$PS = 0$

$y=-3\cos\left(\dfrac{x\pi}{2}\right)+2$
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$};
\draw[very thick,color=red] plot [domain={-360/90}:{360/90},smooth] (\x,{sin(90*\x)});
\end{tikzpicture}$

 

[skeeter]

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

 

[karush]

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

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

so then $\omega=\dfrac{\pi}{2}$

 

[skeeter]

so then $\omega=\dfrac{\pi}{2}$

yes, and $\phi = 0$