Hey, everyone. In this problem, we're asked to evaluate the expression that the secant of the inverse cosine of negative square root of 3 over 2. Now this function might look a little bit complicated to you, but we can still break this down and solve this the way that we already know how. So let's start with that inside function here, the inverse cosine of negative square root of 3 over 2. Now because this is an inverse trigonometric function, remember that we can also think of this as, okay, the cosine of what angle will give me negative square root of 3 over 2? Now also remember, because this is an inverse trigonometric function, we're looking for an angle within a specified interval. Now for the inverse cosine, I only want to work with angles that are between 0 and pi. So, I am looking for an angle between 0 and pi for which my cosine is equal to negative square root of 3 over 2. Now looking at my unit circle, I know that for 5 pi over 6, that will give me a cosine of negative square root of 3 over 2. So that tells me that this inside function is 5 pi over 6. And I am now left to find the secant of that 5 pi over 6. Now from working with these trigonometric functions previously, you may remember that the secant is really just one over the cosine. So this secant of 5 pi over 6 can really be rewritten as one over the cosine of 5 pi over 6. Now we actually just saw what the cosine of 5 pi over 6 was. We know that it's negative square root of 3 over 2. So this is really one over negative square root of 3 over 2. Now whenever I have one over a fraction, I can really just flip that fraction. So this is going to give me negative two over the square root of 3. Now this is technically a correct answer, but whenever we have a radical in our denominator, we really don't want that there. So we want to rationalize this denominator, which we can do by multiplying by square root of 3 over square root of 3. Now that will give me a final answer of negative 2 square root of 3 over 3. And that's my final answer. The secant of the inverse cosine of negative square root of 3 over 2 is negative 2 square root of 3 over 3. Thanks for watching, and let me know if you have questions.
Table of contents
- 0. Fundamental Concepts of Algebra3h 29m
- 1. Equations and Inequalities3h 27m
- 2. Graphs1h 43m
- 3. Functions & Graphs2h 17m
- 4. Polynomial Functions1h 54m
- 5. Rational Functions1h 23m
- 6. Exponential and Logarithmic Functions2h 28m
- 7. Measuring Angles39m
- 8. Trigonometric Functions on Right Triangles2h 5m
- 9. Unit Circle1h 19m
- 10. Graphing Trigonometric Functions1h 19m
- 11. Inverse Trigonometric Functions and Basic Trig Equations1h 41m
- 12. Trigonometric Identities 2h 34m
- 13. Non-Right Triangles1h 38m
- 14. Vectors2h 25m
- 15. Polar Equations2h 5m
- 16. Parametric Equations1h 6m
- 17. Graphing Complex Numbers1h 7m
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- 20. Sequences, Series & Induction1h 15m
- 21. Combinatorics and Probability1h 45m
- 22. Limits & Continuity1h 49m
- 23. Intro to Derivatives & Area Under the Curve2h 9m
11. Inverse Trigonometric Functions and Basic Trig Equations
Evaluate Composite Trig Functions
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