How do you evaluate $cos (2 arctan (- 2))$ $cos(2 arctan (-2))$ ?

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Answer 1 · 2016-05-27T04:31:16

$- \frac{3}{5}$ $-3/5$ .

Let $a = a r c tan (- 2)$ $a = arc tan (-2)$ . Then, $tan a = - 2 < 0$ $tan a = -2 < 0$ . a is in the 2nd quadrant

or in the 4th. So, $cos a = \pm \frac{1}{\sqrt{5}}$ $cos a = +-1/sqrt 5$

The given expression is $cos 2 a = 2 {cos}^{2} a - 1 = \frac{2}{5} - 1 = - \frac{3}{5}$ $cos 2a = 2 cos^2 a - 1 = 2/5-1=-3/5$

How do you evaluate cos(2arctan(−2))cos(2 arctan (-2))?