How do you simplify # tan[cos^-1(-(sqrt3/2))]#?

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Answer 1 · 2018-07-22T00:38:26

#- 1/sqrt 3#.

The cosine value in brackets being negative,

#cos^ (-1)(-sqrt3/2) in [pi/2, pi ]#. So,

#tan cos^ (-1)(-sqrt3/2) #

#= tan (cos^ (-1)cos (pi-pi/6))#

#= tan (pi - pi/6)#

#= -tan (pi/6)#

#=-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) = pi - cos^(-1)(x)#

Using that, your expression can be simplified to

#tan(pi - cos^-1(sqrt(3)/2))#

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

# cos^-1(sqrt(3)/2) = pi/6#

Plugging that into #tan(pi - cos^-1(sqrt(3)/2))# we get

#tan(pi - pi/6) = tan((5pi)/6)#

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

I hope this helps,
Steve