How do you find the exact value of $tan ({cos}^{- 1} (\frac{1}{2}))$ $tan(cos^-1 (1/2))$ ?

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How do you find the exact value of $tan ({cos}^{- 1} (\frac{1}{2}))$ $tan(cos^-1 (1/2))$ ?

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Answer 1 · 2017-02-17T05:46:18

tan (pi/3) = sqrt3
tan (-pi/3) = - sqrt3

Explanation:

${cos}^{- 1} (\frac{1}{2})$ $cos ^-1 (1/2)$ --> $arccos (\frac{1}{2})$ $arccos (1/2)$
Trig table and unit circle -->
$cos x = \frac{1}{2}$ $cos x = 1/2$ --> arc $x = \pm \frac{π}{3}$ $x = +- pi/3$
There are 2 answers for $(0, 2 π)$ $(0, 2pi)$
$tan (\frac{π}{3}) = \sqrt{3}$ $tan (pi/3) = sqrt3$
$tan (- \frac{π}{3}) = - \sqrt{3}$ $tan (-pi/3) = - sqrt3$

Answer 2 · 2017-03-20T20:03:24

see below

Explanation:

From the ${cos}^{- 1} (\frac{1}{2})$ $color(red)(cos^-1(1/2)$ we have the side $a d j a c e n t$ $color(red)(adjacent$ to $∠ θ = 1$ $color(red)(color(red)(angle theta=1$ and the side $h y p o t e n u s e = 2$ $color(red)(hypote n use = 2$ . So this is a $30^{\circ} - 60^{\circ} - 90^{\circ}$ $color(red)(30^@-60^@-90^@$ triangle and $θ = 60^{\circ}$ $color(red)(theta = 60^@$ therefore the $o p p o s i t e$ $color(red)(opposite$ to $∠ θ = \sqrt{3}$ $color(red)(angle theta = sqrt3$ .

Note that from the restrictions for the range of inverse circular functions ${cos}^{- 1} x$ $color(red)(cos^-1 x$ is restricted to quadrants 1 and 2 and since the argument is positive $\frac{1}{2}$ $color(red)(1/2)$ our answer will be in quadrant 1 only.

Hence,
$tan ({cos}^{- 1} (\frac{1}{2})) = tan θ = \frac{o p p o s i t e}{a d j a c e n t} = \frac{\sqrt{3}}{1} = \sqrt{3}$ $color(blue)(tan(cos^-1(1/2))=tan theta=(opposite)/(adjacent)=sqrt3/1=sqrt3$

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How do you find the exact value of $tan ({cos}^{- 1} (\frac{1}{2}))$ $tan(cos^-1 (1/2))$ ?

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