A1. What is $\frac{13}{5} \sin x$, if $\frac{13}{12} \cos x=-1$?
(A) 1
(B) $\frac{5}{13}$
(C) $-\frac{5}{13}$
(D) -1
(E) $\pm 1$
A1. Express $\cos x=-\frac{12}{13}$ and use the relationship $\sin ^{2} x+\cos ^{2} x=1$. Calculate $\sin ^{2} x: \sin ^{2} x=$ $1-\cos ^{2} x=1-\left(-\frac{12}{13}\right)^{2}=\frac{25}{169}$. Since $\sin x= \pm \frac{5}{13}$, the value of the expression $\frac{13}{5} \sin x=\frac{13}{5} \cdot\left( \pm \frac{5}{13}\right)= \pm 1$. The correct answer is $E$.