## Trigonometry (11th Edition) Clone

We first write the equation in the form $y=a \cos [b(x-d)]$. Therefore, $y=2\sin (3x-4)$ becomes $y=2\sin [3(x-\frac{4}{3})]$. Comparing the two equations, $a=2,b=3$ and $d=\frac{4}{3}$. The amplitude is $|a|=|2|=2.$ The period is $\frac{2\pi}{b}=\frac{2\pi}{3}$. The phase shift is $|d|=|\frac{4}{3}|=\frac{4}{3}.$ Therefore, the description in option D is the correct answer.