Suppose angles $A$ and $B$ are in the first quadrant and $\sin(A)=\frac{1}{8}$ and $\sin(B)=$$\frac{12}{13}$.

(a) Find $\cos(A)$ and $\cos(B)$ exactly.
$\cos(A)=$
$\cos(B)=$

(b) Find $\sin(A+B)$ and $\sin(A-B)$ exactly. The exact value of $\sin(A+B)$ is...

A. $\frac{1}{8}\cdot\frac{3}{5}+\frac{4}{5}\cdot\sqrt{1-\frac{1}{8^2}}$
B. $\frac{1}{8}\cdot\frac{4}{5}+\frac{3}{5}\cdot\sqrt{1-\frac{1}{8^2}}$
C. $\frac{1}{8}\cdot\frac{5}{13}+\frac{12}{13}\cdot\sqrt{1-\frac{1}{8^2}}$
D. $\frac{1}{8}\cdot\frac{3}{5}-\frac{4}{5}\cdot\sqrt{1-\frac{1}{8^2}}$
E. $\frac{1}{8}\cdot\frac{12}{13}-\frac{5}{13}\cdot\sqrt{1-\frac{1}{8^2}}$
F. $\frac{1}{8}\cdot\frac{12}{13}+\frac{5}{13}\cdot\sqrt{1-\frac{1}{8^2}}$
G. $\frac{1}{8}\cdot\frac{4}{5}-\frac{3}{5}\cdot\sqrt{1-\frac{1}{8^2}}$
H. $\frac{1}{8}\cdot\frac{5}{13}-\frac{12}{13}\cdot\sqrt{1-\frac{1}{8^2}}$
I. None of the above

The exact value of $\sin(A-B)$ is...

A. $\frac{1}{8}\cdot\frac{3}{5}+\frac{4}{5}\cdot\sqrt{1-\frac{1}{8^2}}$
B. $\frac{1}{8}\cdot\frac{12}{13}-\frac{5}{13}\cdot\sqrt{1-\frac{1}{8^2}}$
C. $\frac{1}{8}\cdot\frac{12}{13}+\frac{5}{13}\cdot\sqrt{1-\frac{1}{8^2}}$
D. $\frac{1}{8}\cdot\frac{5}{13}+\frac{12}{13}\cdot\sqrt{1-\frac{1}{8^2}}$
E. $\frac{1}{8}\cdot\frac{4}{5}-\frac{3}{5}\cdot\sqrt{1-\frac{1}{8^2}}$
F. $\frac{1}{8}\cdot\frac{5}{13}-\frac{12}{13}\cdot\sqrt{1-\frac{1}{8^2}}$
G. $\frac{1}{8}\cdot\frac{4}{5}+\frac{3}{5}\cdot\sqrt{1-\frac{1}{8^2}}$
H. $\frac{1}{8}\cdot\frac{3}{5}-\frac{4}{5}\cdot\sqrt{1-\frac{1}{8^2}}$
I. None of the above