Question

(I think I know it just checking) Which function has a period equal to half the period of the function in y=5cos(3/5x-pi)+4.. the period is 10pi/3
I think its the first one.

1. y=5sin(6/5x-pi)-4
2. y=-5/2sin(6/10x-2pi)+1/2
3. y=-4cos(5/6x-2pi)+5
4. y=2cos(3/10x-pi)-5/2

Answer 1

Answer:

A is your answer

Step-by-step explanation:

Answer 2

Answer:

$y=5sin(\frac{6}{5}x- \pi )-4$

Step-by-step explanation:

The given function is

$y=5cos(\frac{3}{5}x- \pi )+4$

Where $T=\frac{10 \pi}{3}$

Notice half the period is

$\frac{T}{2}=\frac{\frac{10 \pi}{3} }{2}=\frac{10 \pi}{6}=\frac{5 \pi}{3}$

Now, in the first function

$y=5sin(\frac{6}{5}x- \pi )-4$

Notice that this function is in the form: $y=Asin(\omega x + \phi)$

Where $\omega =\frac{6}{5}$, which definition is $\omega = \frac{2 \pi}{T}$

Replacing this value, we have

$\frac{6}{5}=\frac{2 \pi}{T}\\ T=\frac{10 \pi}{6}\\ T=\frac{5 \pi}{3}$

Which means the first function is has half the period of the given function.