How do you simplify $tan [{cos}^{- 1} (- (\frac{\sqrt{3}}{2}))]$ $tan[cos^-1(-(sqrt3/2))]$ ?

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Answer 1 · 2018-07-22T00:38:26

$- \frac{1}{\sqrt{3}}$ $- 1/sqrt 3$ .

The cosine value in brackets being negative,

${cos}^{- 1} (- \frac{\sqrt{3}}{2}) \in [\frac{π}{2}, π]$ $cos^ (-1)(-sqrt3/2) in [pi/2, pi ]$ . So,

${tan cos}^{- 1} (- \frac{\sqrt{3}}{2})$ $tan cos^ (-1)(-sqrt3/2)$

$= tan ({cos}^{- 1} cos (π - \frac{π}{6}))$ $= tan (cos^ (-1)cos (pi-pi/6))$

$= tan (π - \frac{π}{6})$ $= tan (pi - pi/6)$

$= - tan (\frac{π}{6})$ $= -tan (pi/6)$

$= - \frac{1}{\sqrt{3}}$ $=-1/sqrt 3$

Answer 2 · 2018-07-22T00:47:28

-0.5774

There is an Inverse Trigonometric Identity that will help on this webpage
https://brilliant.org/wiki/inverse-trigonometric-identities/

It says that ${cos}^{- 1} (- x) = π - {cos}^{- 1} (x)$ $cos^(-1)(-x) = pi - cos^(-1)(x)$

Using that, your expression can be simplified to

$tan (π - {cos}^{- 1} (\frac{\sqrt{3}}{2}))$ $tan(pi - cos^-1(sqrt(3)/2))$

Do you need to take it further? Taking just part of the above

${cos}^{- 1} (\frac{\sqrt{3}}{2}) = \frac{π}{6}$ $cos^-1(sqrt(3)/2) = pi/6$

Plugging that into $tan (π - {cos}^{- 1} (\frac{\sqrt{3}}{2}))$ $tan(pi - cos^-1(sqrt(3)/2))$ we get

$tan (π - \frac{π}{6}) = tan (\frac{5 π}{6})$ $tan(pi - pi/6) = tan((5pi)/6)$

Using my calculator, the whole thing is equal to -0.5774...

I hope this helps,
Steve

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