Answer
(a) $\frac{3\pi}{2}$, $\frac{5\pi}{2}$
(b) $-\pi$
(c) out of phase.
(d) see graph.
Work Step by Step
(a) Base on the definition of phase $b$ in the standard formula $y=a\cdot sin(kt-b)$, we can identify the
phase for $y_1$ as $b_1=\frac{3\pi}{2}$, and for $y_2$ as $b_2=\frac{5\pi}{2}$
(b) The phase difference between the curves can be found as $b_1-b_2=\frac{3\pi}{2}-\frac{5\pi}{2}=-\pi$
(c) The curves are out of phase because the phase difference is not a multiple of $2\pi$
(d) Both curves are sketched on the same axes as shown in the figure.
