(a) Using an angle-sum or angle-difference identity, rewrite the quantity $8\cos(\frac{x}{2}+\frac{\pi}{4})$ as a sum or difference of two trigonometric functions.


A. $-8\cdot\frac{1}{2}\cos(\frac{x}{2})+8\cdot\frac{\sqrt{3}}{2}\sin(\frac{x}{2})$
B. $-8\cdot\frac{1}{2}\cos(\frac{x}{2})-8\cdot\frac{\sqrt{3}}{2}\sin(\frac{x}{2})$
C. $-8\cdot\frac{\sqrt{2}}{2}\cos(\frac{x}{2})+8\cdot\frac{\sqrt{2}}{2}\sin(\frac{x}{2})$
D. $8\cdot\frac{\sqrt{3}}{2}\cos(\frac{x}{2})-8\cdot\frac{\sqrt{3}}{2}\sin(\frac{x}{2})$
E. $-8\cdot\frac{\sqrt{3}}{2}\cos(\frac{x}{2})-8\cdot\frac{\sqrt{3}}{2}\sin(\frac{x}{2})$
F. $8\cdot\frac{1}{2}\cos(\frac{x}{2})-8\cdot\frac{\sqrt{3}}{2}\sin(\frac{x}{2})$
G. $8\cdot\frac{\sqrt{2}}{2}\cos(\frac{x}{2})-8\cdot\frac{\sqrt{2}}{2}\sin(\frac{x}{2})$
H. $8\cdot\frac{1}{2}\cos(\frac{x}{2})+8\cdot\frac{\sqrt{3}}{2}\sin(\frac{x}{2})$
I. $8\cdot\frac{\sqrt{3}}{2}\cos(\frac{x}{2})-8\cdot\frac{1}{2}\sin(\frac{x}{2})$
J. $-8\cdot\frac{\sqrt{2}}{2}\cos(\frac{x}{2})-8\cdot\frac{\sqrt{2}}{2}\sin(\frac{x}{2})$
K. $8\cdot\frac{\sqrt{3}}{2}\cos(\frac{x}{2})+8\cdot\frac{1}{2}\sin(\frac{x}{2})$
L. $8\cdot\frac{\sqrt{3}}{2}\cos(\frac{x}{2})+8\cdot\frac{\sqrt{3}}{2}\sin(\frac{x}{2})$
M. $8\cdot\frac{\sqrt{2}}{2}\cos(\frac{x}{2})+8\cdot\frac{\sqrt{2}}{2}\sin(\frac{x}{2})$
N. $-8\cdot\frac{\sqrt{3}}{2}\cos(\frac{x}{2})+8\cdot\frac{1}{2}\sin(\frac{x}{2})$
O. $-8\cdot\frac{\sqrt{3}}{2}\cos(\frac{x}{2})-8\cdot\frac{1}{2}\sin(\frac{x}{2})$
P. $-8\cdot\frac{\sqrt{3}}{2}\cos(\frac{x}{2})+8\cdot\frac{\sqrt{3}}{2}\sin(\frac{x}{2})$
Q. None of the above

(b) Suppose $\sin(x)=$$\frac{12}{13}$, and $x$ is between $\frac{\pi}{2}$ and $\pi$. Find $\cos(x)$.
$\cos(x)=$

(c) With $x$ as in part (b), find $8\cos(\frac{x}{2}+\frac{\pi}{4})$ exactly.


A. -$8\cdot\frac{\sqrt{2}}{2}\cdot\sqrt{\frac{1-\frac{5}{13}}{2}}+8\cdot\frac{\sqrt{2}}{2}\cdot\sqrt{\frac{1+\frac{5}{13}}{2}}$
B. -$8\cdot\frac{\sqrt{2}}{2}\cdot\sqrt{\frac{1+\frac{5}{13}}{2}}+8\cdot\frac{\sqrt{2}}{2}\cdot\sqrt{\frac{1-\frac{5}{13}}{2}}$
C. -$8\cdot\frac{\sqrt{2}}{2}\cdot\sqrt{\frac{1-\frac{5}{13}}{2}}-8\cdot\frac{\sqrt{2}}{2}\cdot\sqrt{\frac{1+\frac{5}{13}}{2}}$
D. $8\cdot\frac{\sqrt{2}}{2}\cdot\sqrt{\frac{1+\frac{5}{13}}{2}}-8\cdot\frac{\sqrt{2}}{2}\cdot\sqrt{\frac{1-\frac{5}{13}}{2}}$
E. -$8\cdot\frac{\sqrt{3}}{2}\cdot\sqrt{\frac{1-\frac{5}{13}}{2}}+8\cdot\frac{\sqrt{2}}{2}\cdot\sqrt{\frac{1+\frac{5}{13}}{2}}$
F. -$8\cdot\frac{\sqrt{3}}{2}\cdot\sqrt{\frac{1+\frac{5}{13}}{2}}+8\cdot\frac{\sqrt{2}}{2}\cdot\sqrt{\frac{1-\frac{5}{13}}{2}}$
G. $8\cdot\frac{\sqrt{3}}{2}\cdot\sqrt{\frac{1-\frac{5}{13}}{2}}-8\cdot\frac{\sqrt{2}}{2}\cdot\sqrt{\frac{1+\frac{5}{13}}{2}}$
H. $8\cdot\frac{\sqrt{3}}{2}\cdot\sqrt{\frac{1-\frac{5}{13}}{2}}+8\cdot\frac{\sqrt{2}}{2}\cdot\sqrt{\frac{1+\frac{5}{13}}{2}}$
I. -$8\cdot\frac{\sqrt{2}}{2}\cdot\sqrt{\frac{1+\frac{5}{13}}{2}}-8\cdot\frac{\sqrt{2}}{2}\cdot\sqrt{\frac{1-\frac{5}{13}}{2}}$
J. $8\cdot\frac{\sqrt{2}}{2}\cdot\sqrt{\frac{1-\frac{5}{13}}{2}}+8\cdot\frac{\sqrt{2}}{2}\cdot\sqrt{\frac{1+\frac{5}{13}}{2}}$
K. -$8\cdot\frac{\sqrt{3}}{2}\cdot\sqrt{\frac{1-\frac{5}{13}}{2}}-8\cdot\frac{\sqrt{2}}{2}\cdot\sqrt{\frac{1+\frac{5}{13}}{2}}$
L. $8\cdot\frac{\sqrt{2}}{2}\cdot\sqrt{\frac{1+\frac{5}{13}}{2}}+8\cdot\frac{\sqrt{2}}{2}\cdot\sqrt{\frac{1-\frac{5}{13}}{2}}$
M. $8\cdot\frac{\sqrt{3}}{2}\cdot\sqrt{\frac{1+\frac{5}{13}}{2}}-8\cdot\frac{\sqrt{2}}{2}\cdot\sqrt{\frac{1-\frac{5}{13}}{2}}$
N. -$8\cdot\frac{\sqrt{3}}{2}\cdot\sqrt{\frac{1+\frac{5}{13}}{2}}-8\cdot\frac{\sqrt{2}}{2}\cdot\sqrt{\frac{1-\frac{5}{13}}{2}}$
O. $8\cdot\frac{\sqrt{3}}{2}\cdot\sqrt{\frac{1+\frac{5}{13}}{2}}+8\cdot\frac{\sqrt{2}}{2}\cdot\sqrt{\frac{1-\frac{5}{13}}{2}}$
P. $8\cdot\frac{\sqrt{2}}{2}\cdot\sqrt{\frac{1-\frac{5}{13}}{2}}-8\cdot\frac{\sqrt{2}}{2}\cdot\sqrt{\frac{1+\frac{5}{13}}{2}}$
Q. None of the above