How do you do simplify $sin [2 {tan}^{- 1} (- 2)]$ $sin [2 tan^-1 (-2)]$ ?

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How do you do simplify $sin [2 {tan}^{- 1} (- 2)]$ $sin [2 tan^-1 (-2)]$ ?

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Answer 1 · 2016-04-19T14:20:03

$- \frac{4}{\sqrt{5}}$ $-4/sqrt5$ .

Explanation:

Let $a = {tan}^{- 1} (- 2$ $a = tan^(-1)(-2$ .

$tan a = - 2$ $tan a = -2$ .

As tan a is negative , sin a and cos a have opposite signs.
So, $sin a = \pm \frac{2}{\sqrt{5}}$ $sin a = +-2/sqrt5$ and, correspondingly, $cos a = - \frac{1}{\sqrt{5}} or \frac{1}{\sqrt{5}}$ $cos a = -1/sqrt5 or 1/sqrt5$ .
The given expression is $sin 2 a = 2 sin a cos a = - \frac{4}{5}$ $sin 2a = 2 sin a cos a = -4/5$ .

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How do you do simplify $sin [2 {tan}^{- 1} (- 2)]$ $sin [2 tan^-1 (-2)]$ ?

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