Use the half-angle formulas to evaluate the following exactly.

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

A. $\sqrt{\frac{1-1/2}{2}}$
B. $-\sqrt{\frac{1-\sqrt{2}/2}{2}}$
C. $\sqrt{\frac{1+1/2}{2}}$
D. $-\sqrt{\frac{1+1/2}{2}}$
E. $\sqrt{\frac{1-\sqrt{3}/2}{2}}$
F. $\sqrt{\frac{1+\sqrt{2}/2}{2}}$
G. $-\sqrt{\frac{1+\sqrt{3}/2}{2}}$
H. $-\sqrt{\frac{1+\sqrt{2}/2}{2}}$
I. $\sqrt{\frac{1+\sqrt{3}/2}{2}}$
J. $-\sqrt{\frac{1-1/2}{2}}$
K. $\sqrt{\frac{1-\sqrt{2}/2}{2}}$
L. $-\sqrt{\frac{1-\sqrt{3}/2}{2}}$
M. None of the above

(b) cos($\frac{7\pi}{8}$)=? Careful! Draw a sketch. In which quadrant is $\frac{7\pi}{8}$?

A. $-\sqrt{\frac{1+\sqrt{3}/2}{2}}$
B. $-\sqrt{\frac{1-\sqrt{3}/2}{2}}$
C. $-\sqrt{\frac{1-1/2}{2}}$
D. $\sqrt{\frac{1-\sqrt{3}/2}{2}}$
E. $-\sqrt{\frac{1+1/2}{2}}$
F. $-\sqrt{\frac{1-\sqrt{2}/2}{2}}$
G. $-\sqrt{\frac{1+\sqrt{2}/2}{2}}$
H. $\sqrt{\frac{1-\sqrt{2}/2}{2}}$
I. $\sqrt{\frac{1+1/2}{2}}$
J. $\sqrt{\frac{1-1/2}{2}}$
K. $\sqrt{\frac{1+\sqrt{2}/2}{2}}$
L. $\sqrt{\frac{1+\sqrt{3}/2}{2}}$
M. None of the above