Problem

Find the exact value of each expression. (a) $…

03:36

Problem 67 Hard Difficulty

Find the exact value of each expression.

(a) $ \cot^{-1} (-\sqrt{3}) $
(b) $ \sec^{-1} 2 $

Answer

(a) $\left.\cot ^{-1}(-\sqrt{3})=\frac{5 \pi}{6} \text { because cot } \frac{5 x}{6}=-\sqrt{3} \text { and } \frac{3 x}{6} \text { is in }(0, \pi) \text { (the range of } \cot ^{-1}\right)$.
(b) $\left.\sec ^{-1} 2=\frac{\pi}{3} \text { because sec } \frac{\pi}{3}=2 \text { and } \frac{\pi}{3} \text { is in }\left[0, \frac{\pi}{2}\right) \cup\left[\pi, \frac{3 \pi}{2}\right) \text { (the range of } \sec ^{-1}\right)$.

Johns Hopkins University

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Video Transcript

when you see an expression like this inverse co tangent of negative Square root three. Here's what you want to think. It's asking you to find the angle always within very strict. You're finding the angle. Find the angle whose co tangent is negative Square root three. And you could think of negative square root three as a fraction negative square 3/1. So when finding the angle, remember the range for inverse co tangent. You have to have angles between zero and pie only. And if you have a positive value like positive squarer three you had been, you would be in quadrant one. If you have a negative value, you would be in Quadrant two. We have a negative value, so we're in Quadrant two. Okay, so let's draw a reference triangle and co tangent is adjacent over opposite. So on the adjacent we would have negative square root three. And on the opposite, we would have won at this point. Hopefully, we recognize this reference triangle as a special right triangle. 30 60 90 high pot news would be to, and we have a 30 degree reference angle. So the angle that were interested in the the standard position angle starting from zero ongoing around counterclockwise would be 150 degrees, but we don't want to give our angle in degrees. We want to give it in radiance. 150 degrees is equivalent to five pi over six radiance. Okay, now for the other one. Same idea. When you see in Versi count of two, you want to think of it as find the angle. Who's C can't is to and you can think of to as to over one. Find the angle and remember your outputs for in mercy. Can't have to be in quadrant one or quadrant two between zero and pi over two or between pirate too and pie. And because two is a positive, we're going to be in quadrant one. So we draw reference triangle. We remember that. See Count is high pot news over adjacent so we can put two on high pot news and one on the adjacent and hopefully now we recognize that as our special right triangle with sides 12 in square root three and we would have a 60 degree reference angle. So our answer is 60 degrees, except when you to convert that into radiance. So we get pi over three

Heather Z.

Oregon State University

Related Courses

Calculus 1 / AB

Calculus 2 / BC

Calculus 3

Calculus: Early Transcendentals

Chapter 1

Functions and Models

Section 5

Inverse Functions and Logarithms

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