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Answer 1

Answer on Question #83569 - Math - Trigonometry

Question

What is the value of $\sin \theta$ if $\cos(-\theta) = -\frac{\sqrt{3}}{4}$ , $\sin \theta < 0$ ?

$$- \frac {\sqrt {1 3}}{1 6}$$ $$\frac {\sqrt {1 3}}{1 6}$$ $$- \frac {\sqrt {1 3}}{4}$$ $$\frac {\sqrt {1 3}}{4}$$

Solution

If $\cos (-\theta) = \cos \theta$ then $\cos \theta = -\frac{\sqrt{3}}{4}$ .

Using the basic trigonometric identity $\sin^2\theta +\cos^2\theta = 1;$

$$\sin^ {2} \theta = 1 - \cos^ {2} \theta ;$$ $$\sin \theta = \pm \sqrt {1 - \cos^ {2} \theta}.$$

If $\sin \theta < 0$ then $\sin \theta = -\sqrt{1 - \cos^2\theta}$ ;

$$\sin \theta = - \sqrt {1 - \left(- \frac {\sqrt {3}}{4}\right) ^ {2}};$$ $$\sin \theta = - \sqrt {1 - \frac {3}{1 6}};$$ $$\sin \theta = - \sqrt {\frac {1 6}{1 6} - \frac {3}{1 6}};$$ $$\sin \theta = - \sqrt {\frac {1 6 - 3}{1 6}};$$ $$\sin \theta = - \sqrt {\frac {1 3}{1 6}};$$ $$\sin \theta = - \frac {\sqrt {1 3}}{\sqrt {1 6}};$$ $$\sin \theta = - \frac {\sqrt {13}}{4};$$

Answer: $\sin \theta = -\frac{\sqrt{13}}{4}$.

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