## Trigonometry (11th Edition) Clone

The amplitude is $3$, the period is $2\pi$, there is no vertical translation and the phase shift is $\frac{\pi}{2}$ to the left since $d$ is less than zero.
We first write the equation in the form $y=c+a \sin [b(x-d)]$. Therefore, $y=3\sin (x+\frac{\pi}{2})$ becomes $y=0+3\sin [1(x+\frac{\pi}{2})]$. Comparing the two equations, $a=3,b=1$,c=0 and $d=-\frac{\pi}{2}$. The amplitude is $|a|=|3|=3.$ The period is $\frac{2\pi}{b}=\frac{2\pi}{1}=2\pi$. The vertical translation is $c=0$. The phase shift is $|d|=|-\frac{\pi}{2}|=\frac{\pi}{2}$ Therefore, the amplitude is $3$, the period is $2\pi$, there is no vertical translation and the phase shift is $\frac{\pi}{2}$ to the left since $d$ is less than zero.