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Question: $\frac{5\pi}{6}$ is the principal value of which of the following inverse trigonometric functions?...

5π6\frac{5\pi}{6} is the principal value of which of the following inverse trigonometric functions?

A

sin−112^{-1}\frac{1}{2}

B

cos−112^{-1}\frac{1}{\sqrt{2}}

C

sec−1(−23)^{-1}(-\frac{2}{\sqrt{3}})

D

tan−1(−13)^{-1}(-\frac{1}{\sqrt{3}})

Answer

c. sec−1(−23)^{-1}(-\frac{2}{\sqrt{3}})

Explanation

Solution

To find which inverse trigonometric function has 5π6\frac{5\pi}{6} as its principal value, we need to evaluate the principal value for each given option.

The principal value branches for the inverse trigonometric functions involved are:

  • sin⁡−1x\sin^{-1}x: [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}]
  • cos⁡−1x\cos^{-1}x: [0,π][0, \pi]
  • sec⁡−1x\sec^{-1}x: [0,π]−{π2}[0, \pi] - \{\frac{\pi}{2}\}
  • tan⁡−1x\tan^{-1}x: (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2})

Let's check each option:

a. sin⁡−112\sin^{-1}\frac{1}{2}

We know that sin⁡(π6)=12\sin(\frac{\pi}{6}) = \frac{1}{2}. Since π6∈[−π2,π2]\frac{\pi}{6} \in [-\frac{\pi}{2}, \frac{\pi}{2}], the principal value of sin⁡−112\sin^{-1}\frac{1}{2} is π6\frac{\pi}{6}. π6≠5π6\frac{\pi}{6} \neq \frac{5\pi}{6}.

b. cos⁡−112\cos^{-1}\frac{1}{\sqrt{2}}

We know that cos⁡(π4)=12\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}. Since π4∈[0,π]\frac{\pi}{4} \in [0, \pi], the principal value of cos⁡−112\cos^{-1}\frac{1}{\sqrt{2}} is π4\frac{\pi}{4}. π4≠5π6\frac{\pi}{4} \neq \frac{5\pi}{6}.

c. sec⁡−1(−23)\sec^{-1}(-\frac{2}{\sqrt{3}})

Let y=sec⁡−1(−23)y = \sec^{-1}(-\frac{2}{\sqrt{3}}). Then sec⁡y=−23\sec y = -\frac{2}{\sqrt{3}}. This implies cos⁡y=−32\cos y = -\frac{\sqrt{3}}{2}. We know that cos⁡(π6)=32\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}. Since cos⁡y\cos y is negative and yy must be in the principal value branch [0,π]−{π2}[0, \pi] - \{\frac{\pi}{2}\}, yy must be in the second quadrant. Therefore, y=π−π6=6π−π6=5π6y = \pi - \frac{\pi}{6} = \frac{6\pi - \pi}{6} = \frac{5\pi}{6}. Since 5π6∈[0,π]−{π2}\frac{5\pi}{6} \in [0, \pi] - \{\frac{\pi}{2}\}, this is the principal value. 5π6=5π6\frac{5\pi}{6} = \frac{5\pi}{6}.

d. tan⁡−1(−13)\tan^{-1}(-\frac{1}{\sqrt{3}})

We know that tan⁡(π6)=13\tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}. Since tan⁡y\tan y is negative and yy must be in the principal value branch (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}), yy must be in the fourth quadrant. Therefore, tan⁡−1(−13)=−π6\tan^{-1}(-\frac{1}{\sqrt{3}}) = -\frac{\pi}{6}. −π6≠5π6-\frac{\pi}{6} \neq \frac{5\pi}{6}.

Based on the evaluation, only option c yields 5π6\frac{5\pi}{6} as its principal value.