How do you solve and find the value of ${cos}^{- 1} ({sin}^{- 1} 90)$ $cos^-1(sin^-1 90)$ ?

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How do you solve and find the value of ${cos}^{- 1} ({sin}^{- 1} 90)$ $cos^-1(sin^-1 90)$ ?

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Answer 1 · 2017-11-08T03:10:01

One cannot evaluate ${cos}^{- 1} ({sin}^{- 1} 90)$ $cos^-1(sin^-1 90)$ , because $90$ $90$ is outside the domain for the inverse sine function:

$θ = {sin}^{- 1} (x); - 1 \leq x \leq 1$ $theta = sin^-1(x); -1 <= x <= 1$

Also, please observe that the inverse sine function returns an angle with the units of either degrees or radians; this is incompatible with the argument of the inverse cosine function, which requires a unitless quantity.

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How do you solve and find the value of ${cos}^{- 1} ({sin}^{- 1} 90)$ $cos^-1(sin^-1 90)$ ?

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How do you solve and find the value of cos−1(sin−190)cos^-1(sin^-1 90)?

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How do you solve and find the value of ${cos}^{- 1} ({sin}^{- 1} 90)$ $cos^-1(sin^-1 90)$ ?