Answer:
The [tex]amplitude=|A|=\frac{1}{2}[/tex], [tex]period=3[/tex] and [tex]\text{phase shift}=\frac{2\pi}{3}[/tex].
Step-by-step explanation:
The given function is
[tex]f(t)=-\frac{1}{2}\sin (3t-2\pi)[/tex] .... (1)
The general form of sine function is
[tex]F(t)=A\sin (Bt-C)+D[/tex] ....(2)
where, |A| is the amplitude, B is period, D is the vertical shift (up or down), and C/B is used to find the phase shift.
On comparing (1) and (2), we get
[tex]A=-\frac{1}{2}[/tex]
[tex]B=3[/tex]
[tex]C=2\pi[/tex]
[tex]D=0[/tex]
So,
[tex]amplitude=|A|=\frac{1}{2}[/tex]
[tex]period=3[/tex]
[tex]\text{phase shift}=\frac{2\pi}{3}[/tex]
Therefore [tex]amplitude=|A|=\frac{1}{2}[/tex], [tex]period=3[/tex] and [tex]\text{phase shift}=\frac{2\pi}{3}[/tex].
