How do you calculate $tan (2 arcsin (- \frac{8}{17}))$ $tan(2arcsin(-8/17))$ ?

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How do you calculate $tan (2 arcsin (- \frac{8}{17}))$ $tan(2arcsin(-8/17))$ ?

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Answer 1 · 2016-06-21T12:20:47

$\pm \frac{240}{161}$ $+-240/161$ .

Explanation:

Let $a = a r c sin (- \frac{8}{17})$ $a = arc sin (-8/17)$ . Then $sin a = - \frac{8}{17} < 0 .$ $sin a = -8/17<0.$ .

So, a is in either 3rd quadrant or in the 4th. Accordingly,

$cos a = \pm \sqrt{1 - \frac{8^{2}}{17^{2}}} = \pm \frac{15}{17} and tan a = \pm \frac{8}{15}$ $cos a = +-sqrt(1-8^2/17^2)=+- 15/17 and tan a = +-8/15$ .

Now, the given expression is

$tan 2 a = \frac{2 tan a}{1 - {tan}^{2} a}$ $tan 2a = (2 tan a)/(1-tan^2a)$

$= \pm \frac{\frac{16}{15}}{1 - \frac{8^{2}}{15^{2}}}$ $=+-(16/15)/(1-8^2/15^2)$

$= \pm \frac{240}{161}$ $=+-240/161$ ..

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