Answer
$Amplitude = |\frac{4}{3}| = \frac{4}{3}$
$Period = \frac{2\pi}{3}$
$Horizontal\ shift = (–\frac{\frac{\pi}{2}}{3}) = -\frac{\pi}{6}$
$Phase = \frac{\pi}{2}$
Work Step by Step
If C is any real number and $B> 0$, then the graphs of $y = A\sin(Bx+C)$ and $y = A\cos (Bx+C)$ will have
$Amplitude = |A|$
$Period = \frac{2\pi}{B}$
$Horizontal\ shift = –\frac{C}{B}$
$Phase = C$
so for $y = \frac{4}{3}\cos (3x + \frac{\pi}{2})$
$Amplitude = |\frac{4}{3}| = \frac{4}{3}$
$Period = \frac{2\pi}{3}$
$Horizontal\ shift = (–\frac{\frac{\pi}{2}}{3}) = -\frac{\pi}{6}$
$Phase = \frac{\pi}{2}$
