Answer :
Answer:
[tex]y=5sin(\frac{6}{5}x- \pi )-4[/tex]
Step-by-step explanation:
The given function is
[tex]y=5cos(\frac{3}{5}x- \pi )+4[/tex]
Where [tex]T=\frac{10 \pi}{3}[/tex]
Notice half the period is
[tex]\frac{T}{2}=\frac{\frac{10 \pi}{3} }{2}=\frac{10 \pi}{6}=\frac{5 \pi}{3}[/tex]
Now, in the first function
[tex]y=5sin(\frac{6}{5}x- \pi )-4[/tex]
Notice that this function is in the form: [tex]y=Asin(\omega x + \phi)[/tex]
Where [tex]\omega =\frac{6}{5}[/tex], which definition is [tex]\omega = \frac{2 \pi}{T}[/tex]
Replacing this value, we have
[tex]\frac{6}{5}=\frac{2 \pi}{T}\\ T=\frac{10 \pi}{6}\\ T=\frac{5 \pi}{3}[/tex]
Which means the first function is has half the period of the given function.