To find the exact values of the inverse sine, cosine, and tangent of \( \sqrt{3} \), let's evaluate each function separately.
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Finding \( \sin^{-1}(\sqrt{3}) \): The range of the \( \sin^{-1}(x) \) function is \([- \frac{\pi}{2}, \frac{\pi}{2}]\) and the sine function only takes values in the range \([-1, 1]\). Since \( \sqrt{3} \) is approximately \( 1.732 \) (which is greater than 1), it is not a valid input for \( \sin^{-1} \). Therefore: \[ \sin^{-1}(\sqrt{3}) = \text{The exact value cannot be determined.} \]
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Finding \( \cos^{-1}(\sqrt{3}) \): The range of the \( \cos^{-1}(x) \) function is \([0, \pi]\) and the cosine function takes values from \([-1, 1]\). Since \( \sqrt{3} \) is also greater than 1, it is not a valid input for \( \cos^{-1} \). Therefore: \[ \cos^{-1}(\sqrt{3}) = \text{The exact value cannot be determined.} \]
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Finding \( \tan^{-1}(\sqrt{3}) \): The range of the \( \tan^{-1}(x) \) function is \((- \frac{\pi}{2}, \frac{\pi}{2})\), and it can take any real number as input. The tangent of \( \frac{\pi}{3} \) equals \( \sqrt{3} \). Thus: \[ \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}. \]
In summary, the results are:
- \( \sin^{-1}(\sqrt{3}) = \text{The exact value cannot be determined.} \)
- \( \cos^{-1}(\sqrt{3}) = \text{The exact value cannot be determined.} \)
- \( \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}. \)