Question

Pre-Cal - Someone please help! (Image Attached)

Answer 1

Note that $\displaystyle{ \sin30^{\circ}= \frac{1}{2}$ and $\displaystyle{ \cos30^{\circ}= \frac{ \sqrt{3}}{2}$.

From the unit circle, we can check that $\sin30^{\circ}$ and $\sin(-30)^{\circ}$ have the same value, but opposite signs, that is:

$\sin(-30)^{\circ}=-\sin(30)^{\circ}=-\frac{1}{2}$. 


Thus, 
$\displaystyle{ \cos(x-y)= \cos(-30^{\circ}-60^{\circ})=cos(-90^{\circ})$.


Note that $cos(-90^{\circ})$ is the x-coordinate of the lowest point on the unit circle, that is (0, -1). Thus $cos(-90^{\circ})=0$


Answer: 0