inverse trigonometric functions with plus minus sign

Question

$$ \cos\left(\theta_{} \right) = \pm \frac{ 1 }{ \sqrt{ 3 } } $$

$$ \theta_{} = \pm \cos^{-1} \left( \pm \frac{ 1 }{ \sqrt{ 3 } } \right) $$

I can't get why the leftmost $~ \pm ~$ of the right term of the second equation should be needed?

I thought as

$$ \cos\left(\theta_{} \right) = \pm \frac{ 1 }{ \sqrt{ 3 } } ~~ $$

is held,

$$ \theta_{} = \cos^{-1} \left( \pm \frac{ 1 }{ \sqrt{ 3 } } \right) $$

must be held.

misprint?

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Answer 1

$$ \cos\left(\theta_{} \right) = \pm \frac{ 1 }{ \sqrt{ 3 } } $$

$$ \theta_{} ~~ \leftarrow~~ \text{constant angle } $$

As this argument is thrown to the cosine function ,the 2 way return values exist and it are $~ \pm \frac{ 1 }{ \sqrt{ 3 } } ~$

$$ \theta_{} = \cos^{-1} \left( \pm \frac{ 1 }{ \sqrt{ 3 } } \right) $$

$$ \cos\left(\cos^{-1} \left( \pm \frac{ 1 }{ \sqrt{ 3 } } \right) \right) = \pm \frac{ 1 }{ \sqrt{ 3 } } $$

By the way,

$$ \theta_{} = - \cos^{-1} \left( \pm \frac{ 1 }{ \sqrt{ 3 } } \right) $$

$$ \cos\left(-\cos^{-1} \left( \pm \frac{ 1 }{ \sqrt{ 3 } } \right) \right) $$

$$ = \cos\left(\cos^{-1} \left( \pm \frac{ 1 }{ \sqrt{ 3 } } \right) \right) = \pm \frac{ 1 }{ \sqrt{ 3 } } $$

$$ \therefore ~~ \theta_{} = \pm \cos^{-1} \left( \pm \frac{ 1 }{ \sqrt{ 3 } } \right) \ni \cos\left(\theta_{} \right) = \pm \frac{ 1 }{ \sqrt{ 3 } } $$