Question

cos∝= √3/2 and 270°<∝<360°. Find: a) sin∝, b) tg∝, c) cotg∝

Answer 1

Answer on Question #63371-Math-Trigonometry

$\cos \alpha = \sqrt{3} / 2$ and $270{}^{\circ} < \alpha < 360{}^{\circ}$. Find: a) $\sin \alpha$, b) $\tan \alpha$, c) $\cot \alpha$

Solution.

a) $\sin \alpha$

Find out the basic trigonometric identities

$$\cos^2 \alpha + \sin^2 \alpha = 1,$$ $$\sin^2 \alpha = 1 - \cos^2 \alpha,$$ $$\sin \alpha = \pm \sqrt{1 - \cos^2 \alpha},$$ $$270{}^{\circ} < \alpha < 360{}^{\circ}, \alpha \in \mathrm{IV}, \sin \alpha < 0$$ $$\sin \alpha = -\sqrt{1 - \cos^2 \alpha}, = -\sqrt{1 - \left(\frac{\sqrt{3}}{2}\right)^2} = -\sqrt{1 - \frac{3}{4}} = -\sqrt{\frac{1}{4}} = -\frac{1}{2}$$

Answer: $-\frac{1}{2}$

b) $\tan \alpha$

$$\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = -\frac{1}{2} \cdot \frac{\sqrt{3}}{2} = -\frac{1}{\sqrt{3}}.$$

Answer: $-\frac{1}{\sqrt{3}}$

Find: c) $\cot \alpha$

$$\cot \alpha = \frac{1}{\tan \alpha} = -\frac{1}{\frac{1}{\sqrt{3}}} = -\sqrt{3}.$$

Answer: $-\sqrt{3}$

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