Use the angle sum and difference formulas to evaluate the following exactly. Note that there may be more than one way to evaluate each.

(a) $\cos(\frac{7\pi}{12})$=?

A. $\frac{\sqrt{6}-\sqrt{2}}{4}$
B. $\frac{-\sqrt{2}-\sqrt{3}}{4}$
C. $\frac{\sqrt{3}-\sqrt{2}}{4}$
D. $\frac{\sqrt{2}+\sqrt{6}}{4}$
E. $\frac{\sqrt{6}-\sqrt{3}}{4}$
F. $\frac{\sqrt{2}-\sqrt{3}}{4}$
G. $\frac{\sqrt{3}+\sqrt{6}}{4}$
H. $\frac{\sqrt{2}+\sqrt{3}}{4}$
I. $\frac{\sqrt{3}-\sqrt{6}}{4}$
J. $\frac{\sqrt{2}-\sqrt{6}}{4}$
K. $\frac{-\sqrt{2}-\sqrt{6}}{4}$
L. $\frac{-\sqrt{3}-\sqrt{6}}{4}$
M. None of the above

(b) $\sin(\frac{17\pi}{12})$=?

A. $\frac{\sqrt{2}-\sqrt{6}}{4}$
B. $\frac{\sqrt{3}-\sqrt{6}}{4}$
C. $\frac{-\sqrt{2}-\sqrt{3}}{4}$
D. $\frac{\sqrt{6}-\sqrt{2}}{4}$
E. $\frac{-\sqrt{2}-\sqrt{6}}{4}$
F. $\frac{\sqrt{2}+\sqrt{3}}{4}$
G. $\frac{\sqrt{3}+\sqrt{6}}{4}$
H. $\frac{\sqrt{2}-\sqrt{3}}{4}$
I. $\frac{-\sqrt{3}-\sqrt{6}}{4}$
J. $\frac{\sqrt{2}+\sqrt{6}}{4}$
K. $\frac{\sqrt{6}-\sqrt{3}}{4}$
L. $\frac{\sqrt{3}-\sqrt{2}}{4}$
M. None of the above