How do you evaluate $cos ({tan}^{- 1} (\frac{1}{2}))$ $cos(tan^-1(1/2))$ without a calculator?

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Answer 1 · 2016-09-23T04:04:50

$\pm \frac{2}{\sqrt{5}}$ $+-2/sqrt5$

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

$H = \sqrt{2^{2} + 1^{2}} = \sqrt{5}$ $H=sqrt(2^2+1^2)=sqrt5$

$cos ({tan}^{- 1} (\frac{1}{2})) = cos a = \frac{2}{H} = \frac{2}{\sqrt{5}}$ $cos(tan^-1(1/2))=cos a = 2/H=2/sqrt5$

As tangent is positive in both the first and the third quadrants,
$\Rightarrow cos a = \pm \frac{2}{\sqrt{5}}$ $=> cosa=+-2/sqrt5$

positive sign for $a$ $a$ in the first quadrant and negative sign for $a$ $a$ in the third quadrant.

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