Chapter 13, Trigonometric Ratios and Functions - 13.4 Evaluate Inverse Trigonometric Functions - 13.4 Exercises - Quiz for Lessons 13.3-13.4 - Page 880: 5

$-\dfrac{\sqrt 3}{2}$

Here, we have $\cos 150^{\circ}$ This angle lies in the Second Quadrant. Here, $\theta'=\pi-\theta=180^{\circ}-150^{\circ}=30^{\circ}$ Now, $\cos \theta'=\cos 30^{\circ}=\dfrac{\sqrt 3}{2}$ The function cosine is negative in Quadrant $II$. Thus, $\cos \theta= -\cos \theta'=-\dfrac{\sqrt 3}{2}$