Question

Find the exact value of the inverse sine, cosine, and tangent of √3
, if possible.

sin^−1(√3)
=



π
π/2
π/4
π/3
−π/6
−π/3
The exact value cannot be determined.




cos^−1(√3)
=

π/6
π/4
−π/6
π
π/3
−π/2
The exact value cannot be determined.




tan^−1(√3)
=

−π
−π/2
−π/3
π/3
π/6
π/4
The exact value cannot be determined.

Answer 1

To find the exact values of the inverse sine, cosine, and tangent of \( \sqrt{3} \), let's evaluate each function separately.

1. **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:
\[
\sin^{-1}(\sqrt{3}) = \text{The exact value cannot be determined.}
\]

2. **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:
\[
\cos^{-1}(\sqrt{3}) = \text{The exact value cannot be determined.}
\]

3. **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:
1. \( \sin^{-1}(\sqrt{3}) = \text{The exact value cannot be determined.} \)
2. \( \cos^{-1}(\sqrt{3}) = \text{The exact value cannot be determined.} \)
3. \( \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}. \)