Question

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Answer 1

Answer on Question #83568 – Math – Trigonometry

Question

$$\tan \theta = \sqrt{15}/15$$ $$\sin(-\theta) = -1/4$$

What is the value of $\cos(\theta)$?

$$-\frac{\sqrt{15}}{60}$$ $$-\frac{\sqrt{15}}{4}$$ $$\frac{\sqrt{15}}{60}$$ $$\frac{\sqrt{15}}{4}$$

Answer

INSTRUCTIONS:

$$\sin(-\theta) = -\sin \theta$$ $$-\sin \theta = -\frac{1}{4}$$ $$\sin \theta = 1/4$$

FORMULA:

$$\sin^2 \theta + \cos^2 \theta = 1$$ $$\tan \theta = \frac{\sin \theta}{\cos \theta}$$

SOLUTION:

$$\cos \theta = \pm \sqrt{1 - \sin^2 \theta}$$ $$\cos \theta = \pm \sqrt {1 - \left(\frac {1}{16}\right)}$$ $$\cos \theta = \pm \frac {\sqrt {15}}{4}$$

NOW, $\sin \theta = 1/4$ , if, $\cos \theta = -\sqrt{15}/4$ , $\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{1}{4}}{\frac{\sqrt{15}}{4}} = -\sqrt{15}/15$ , which is not given in the problem...

SO, $\cos \theta = \sqrt{15} / 4,$

$$\sin \theta = 1/4$$ $$\tan \theta = \frac {\frac {1}{4}}{\frac {\sqrt {15}}{4}} = \frac {1}{\sqrt {15}} = \frac {\sqrt {15}}{\sqrt {15} * \sqrt {15}} = \frac {\sqrt {15}}{15} \text{ (cross checking with the problem)}$$

SO, our answer is $\cos \theta = \frac{\sqrt{15}}{4}$ .

Answer: $\cos \theta = \frac{\sqrt{15}}{4}$ .

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