Answer
(a) $\frac{\pi}{2}$ and $\frac{\pi}{2}$
(b) $0$
(c) In phase.
(d) see graph.
Work Step by Step
(a) Based on the definition of phase $b$ in a standard equation $y=a\cdot sin(kt-b)$,
we can identify the phase for $y_1$ as $b_1=\frac{\pi}{2}$ and for $y_2$ as $b_2=\frac{\pi}{2}$
(b) The phase difference between the curves can be found as $b_1-b_2=0$
(c) The curves are in phase because the phase difference is zero.
(d) We can sketch both curves on the same axes as shown in the figure.
