Question

Given tan⁡u=-1/3 and sin⁡〖u<0,〗 find sin⁡(u/2). (Assume 0≤u<2π.)



-√((10+3√10)/20) b. √((10-3√10)/20) c. √((10+3√10)/20)

d. -√((10-3√10)/20) e. None of these

Answer 1

Answer on Question #74960 – Math / Trigonometry

Given $\tan^{[n]}u = -1/3$ and $\sin^{[n]}u \in [u < 0, \text{一}]$ find $\sin^{[n]}u(u/2)$. (Assume $0 \leq u < 2\pi$.)

Solution

$$\tan^2 u + 1 = \frac{1}{\cos^2 u}$$ $$\cos^2 u = \frac{1}{\tan^2 u + 1} = \frac{1}{\frac{1}{9} + 1} = \frac{9}{10}$$ $$\cos u = \frac{3}{\sqrt{10}}$$

We will use the formula:

$$\cos u = 1 - 2 \sin^2 \frac{u}{2}$$ $$2 \sin^2 \frac{u}{2} = 1 - \cos u$$ $$\sin^2 \frac{u}{2} = \frac{1 - \cos u}{2} = \frac{1 - \frac{3}{\sqrt{10}}}{2} = \frac{\sqrt{10} - 3}{2\sqrt{10}} = \frac{10 - 3\sqrt{10}}{20}$$ $$\sin \frac{u}{2} = \pm \sqrt{\frac{10 - 3\sqrt{10}}{20}}$$

If $\sin u < 0$ then $\sin(u/2) < 0$

So,

$$\sin \frac{u}{2} = -\sqrt{\frac{10 - 3\sqrt{10}}{20}}$$

Answer:

$$\sin \frac{u}{2} = -\sqrt{\frac{10 - 3\sqrt{10}}{20}}$$

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