## Trigonometry (11th Edition) Clone

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Convert the angle measure to degrees to obtain: $=-\frac{7\pi}{6} \cdot \frac{180^o}{\pi} = -7(30^o)=-210^o$ Thus, $\sin{(-\frac{7\pi}{6})} = \sin{(-210^o)}$ $-210^o$ is co-terminal with $-210^o+360^o=150^o$. $150^o$ is in Quadrant II so its reference angle is $=180^o-150^o=30^o$. Note that the sine function is positive in Quadrant II.