## Calculus with Applications (10th Edition)

$sin 150^\circ =\frac{1}{2}$ $sec 150^\circ =\frac{-2}{\sqrt 3}$ $cot150^\circ =-\sqrt 3$
From the table we have the following information: $cos 150^\circ=-\frac{\sqrt 3}{2}$ $tan 150^\circ=\frac{-\sqrt 3}{3}$ $csc150^\circ =2$ then, using the elementary trigonometric identities we obtain: $sin 150^\circ =\frac{1}{2}$ $sec 150^\circ =\frac{-2}{\sqrt 3}$ $cot150^\circ =\frac{1}{\tan 150^\circ}=-\sqrt 3$