Chapter 6 - Review Exercises - Page 291: 15

$\theta=-60^o$

Work Step by Step

RECALL: $y=\arcsin{(x)} \longrightarrow \sin{y}=x$, $y$ is in the interval $[90^o, 90^o]$ Thus, $\theta=\arccos{(-\frac{\sqrt3}{2})}$ implies that $\cos{\theta}=-\frac{\sqrt3}{2}$. Note that $\sin{(60^o)}=\frac{\sqrt3}{2}$. Since sine is an odd function, then $\sin{(-x)}=-\sin{x}$. This means that $\sin{(-60^o)}=-\sin{(60^o)}=-\frac{\sqrt3}{2}$ Therefore, $\theta=\arcsin{(-\frac{\sqrt3}{2})}\longrightarrow \theta=-60^o$

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