Question

How do I find the period for d=9cos(pi/2 t)?
Please explain

Answer 1

\[\Large d = 9\cos\left(\frac{\pi}{2}t\right)\]
is the same as
\[\Large d = 9\cos\left(\frac{\pi}{2}t-0\right)+0\]
Agreed? Or no?

Answer 2

let me know where you're stuck @ohohaye

Answer 3

I'm confused. Can you please explain.

Answer 4

I know that the answer is 4 but, I don't know how to get that answer.

Answer 5

I used the idea that

x+0 = x
x-0 = x

x can be any real number

Answer 6

the reason I'm doing it is to get the equation into a certain form

Answer 7

Are you trying to make the equation to equal 0?

Answer 8

no, I'm trying to make it fit the form
\[\Large y = A*\cos(Bt - C) + D\]

Answer 9

notice how
\[\Large d = 9\cos\left(\frac{\pi}{2}t-0\right)+0\]
is in the form
\[\Large y = A*\cos(Bt - C) + D\]

where

A = 9
B = pi/2
C = 0
D = 0

do you see how I'm getting this?

Answer 10

Yeah, I see that now

Answer 11

@jim_thompson5910

Answer 12

To find the period T, we now use the formula

T = 2pi/B

Answer 13

Plug in B = pi/2


T = 2pi/B

T = 2pi/(pi/2)

T = 2pi*(2/pi)

T = 2*2

T = 4


so the period is 4.