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

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

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Answer 1 · 2016-11-25T15:14:28

-4/5

Explanation:

Consider the arc T in Quadrant 4, representing by the right triangle
OBM, with:
tan T = -2 --> OB = 1 ; BM = -2
Consequently, the radius hypotenuse $O M = \sqrt{1 + 4} = \sqrt{5}$ $OM = sqrt(1 + 4) = sqrt5$
In this right triangle OBM,
$sin T = \frac{1}{\sqrt{5}}$ $sin T = 1/sqrt5$ and $cos T = - \frac{2}{\sqrt{5}}$ $cos T = -2/sqrt5$

$sin ((2 {tan}^{- 2} (- 2)) \to sin (2 arctan (- 2)) = sin (2 T)$ $sin ((2tan^-2(-2)) -> sin (2arctan (-2)) = sin (2T)$
Use trig identity: sin 2a = 2sin a.cos a -->

$sin 2 T = 2 sin T . cos T = 2 (\frac{1}{\sqrt{5}}) (- \frac{2}{\sqrt{5}}) = - \frac{4}{5}$ $sin 2T = 2sin T.cos T = 2(1/sqrt5)(-2/sqrt5) = -4/5$

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

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