Answer
a) $-1/2$
b) $-\sqrt2/2$
c) $\sqrt3$
d) $-1$
Work Step by Step
a) sin$(7\pi/6)$ is also equivalent to sin$(\pi)$+sin$(\pi/6)$, but it is not in Quadrant I but in III, so the result is negative. sin$(\pi)=0$ and sin$(\pi/6)=1/2$ Thus sin$(7\pi/6)=-1/2$
b) cos$(13\pi/4)$ is also equivalent to cos$(2\pi)$+cos$(\pi)$+cos$(\pi/4)$.
cos$(2\pi)$ means it will make one full lap on the circle, so no it's unsignificant.
cos$(\pi)$ means it will make half lap and the additional cos$(\pi/4)$ means it will end up in Quadrant III, so the result will be the negative of cos$(\pi/4)=\sqrt2/2$. Thus, cos$(13\pi/4)=-\sqrt2/2$
c) By going $-5\pi/3$, it will only need $-\pi/3$ extra to make a full lap around the circle. Thus, tan$(-5\pi/3)=$ tan$(\pi/3)=\sqrt3$
d) csc$(3\pi/2)$=$\frac{1}{sin(3\pi/2)}=\frac{1}{-1}=-1$
