# Trigonometry-7.8.1 Amp, Period, PS, VS of 3cos(\pi x-2)+5

#### karush

##### Well-known member
Find amplitude, period, PS, VS. graph 2 periods of
$y=3\cos(\pi x-2)+5$

ok I think these are the plug ins we use
$Y=A\cos\left[\omega\left(x-\dfrac{x \phi}{\omega} \right)\right]+B$
or
$A\cos\left(\omega x-\phi\right)+B$
A=amplitude B=VS or veritical shift
$T = \dfrac{2\pi}{\omega-\phi}$
$PS = 0$ assumed here

ok just want to see if I have these plug in eq right, different books use different symbols

#### topsquark

##### Well-known member
MHB Math Helper
Find amplitude, period, PS, VS. graph 2 periods of
$y=3\cos(\pi x-2)+5$

ok I think these are the plug ins we use
$Y=A\cos\left[\omega\left(x-\dfrac{x \phi}{\omega} \right)\right]+B$
or
$A\cos\left(\omega x-\phi\right)+B$
A=amplitude B=VS or veritical shift
$T = \dfrac{2\pi}{\omega-\phi}$
$PS = 0$ assumed here

ok just want to see if I have these plug in eq right, different books use different symbols
Use $$\displaystyle Y = A\cos\left(\omega x-\phi\right)+B$$ or $$\displaystyle Y=A\cos\left[\omega\left(x-\dfrac{\phi}{\omega} \right)\right]+B$$. (You had one too many x's in your first equation.)

-Dan

#### karush

##### Well-known member
$\displaystyle Y=A\cos\left[\omega\left(x-\dfrac{\phi}{\omega} \right)\right]+B$
then for $y=3\cos(\pi x-2)+5$
$A=3 \quad \omega=\pi \quad \phi=2 \quad B=5$
before the plug...
where $T=\dfrac{2\pi}{\omega}$ and $PS=\dfrac{\phi}{\omega}$

#### topsquark

##### Well-known member
MHB Math Helper
$\displaystyle Y=A\cos\left[\omega\left(x-\dfrac{\phi}{\omega} \right)\right]+B$
then for $y=3\cos(\pi x-2)+5$
$A=3 \quad \omega=\pi \quad \phi=2 \quad B=5$
before the plug...
where $T=\dfrac{2\pi}{\omega}$ and $PS=\dfrac{\phi}{\omega}$
Yup.

-Dan

#### karush

##### Well-known member

before the plug...
where $T=\dfrac{2\pi}{\omega}$ and $PS=\dfrac{\phi}{\omega}$
so then
$T=\dfrac{2\pi}{\pi}=2$ and $PS=\dfrac{2}{\pi}$
kinda ??? on PS
So T is Period?

#### topsquark

##### Well-known member
MHB Math Helper

before the plug...
where $T=\dfrac{2\pi}{\omega}$ and $PS=\dfrac{\phi}{\omega}$
so then
$T=\dfrac{2\pi}{\pi}=2$ and $PS=\dfrac{2}{\pi}$
kinda ??? on PS
So T is Period?
Yes. You have it right.

-Dan