<meta http-equiv="refresh" content="1; url=/nojavascript/"> Composition of Inverse Reciprocal Trig Functions ( Read ) | Trigonometry | CK-12 Foundation
Dismiss
Skip Navigation
You are viewing an older version of this Concept. Go to the latest version.

Composition of Inverse Reciprocal Trig Functions

%
Progress
Practice Composition of Inverse Reciprocal Trig Functions
Practice
Progress
%
Practice Now
Composition of Inverse Reciprocal Trig Functions

Composing functions involves applying one function and then applying another function afterward. In the case of inverse reciprocal functions, you could create compositions of functions such as sec^{-1} , csc^{-1} , and cot^{-1} .

Consider the following problem:

\csc (\cot^{-1} \sqrt{3})

Can you solve this problem?

Keep reading, and at the conclusion of this Concept, you'll be able to do so.

Watch This

Hard Trig Inverse Composition Problems

Guidance

Just as you can apply one function and then another whenever you'd like, you can do the same with inverse reciprocal trig functions. This process is called composition. Here we'll explore some examples of composition for these inverse reciprocal trig functions by doing some problems.

Example A

Without a calculator, find \cos \left ( \cot^{-1} \sqrt{3} \right ) .

Solution:

First, find \cot^{-1} \sqrt{3} , which is also \tan^{-1}\frac{\sqrt{3}}{3} . This is \frac{\pi}{6} . Now, find \cos \frac{\pi}{6} , which is \frac{\sqrt{3}}{2} . So, our answer is \frac{\sqrt{3}}{2} .

Example B

Without a calculator, find \sec^{-1} \left ( \csc \frac{\pi}{3} \right ) .

Solution: First, \csc \frac{\pi}{3} = \frac{1}{\sin \frac{\pi}{3}} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3} . Then \sec^{-1} \frac{2\sqrt{3}}{3} = \cos^{-1} \frac{\sqrt{3}}{2} = \frac{\pi}{6} .

Example C

Evaluate \cos \left ( \sin^{-1}  \frac{3}{5}  \right ) .

Solution: Even though this problem is not a critical value, it can still be done without a calculator. Recall that sine is the opposite side over the hypotenuse of a triangle. So, 3 is the opposite side and 5 is the hypotenuse. This is a Pythagorean Triple, and thus, the adjacent side is 4. To continue, let \theta = \sin^{-1}  \frac{3}{5} or \sin \theta = \frac{3}{5} , which means \theta is in the Quadrant 1 (from our restricted domain, it cannot also be in Quadrant II). Substituting in \theta we get \cos \left ( \sin^{-1}  \frac{3}{5}  \right ) = \cos \theta and \cos \theta = \frac{4}{5} .

Vocabulary

Inverse Function: An inverse function is a function that undoes another function.

Reciprocal Function: A reciprocal function is a function that when multiplied by the original function gives the number 1 as a result.

Guided Practice

1. Find the exact value of \csc \left ( \cos^{-1}  \frac{\sqrt{3}}{2}  \right ) without a calculator, over its restricted domains.

2. Find the exact value of \sec^{-1} ( \tan (\cot^{-1} 1)) without a calculator, over its restricted domains.

3. Find the exact value of \tan^{-1} \left ( \cos \frac{\pi}{2} \right ) without a calculator, over its restricted domains.

Solutions:

1. \csc \left ( \cos^{-1}  \frac{\sqrt{3}}{2}  \right ) = \csc \frac{\pi}{6} = 2

2. \sec^{-1} (\tan (\cot^{-1} 1)) = \sec^{-1} \left ( \tan  \frac{\pi}{4}  \right ) = \sec^{-1} 1 = 0

3. \tan^{-1} \left ( \cos \frac{\pi}{2} \right ) = \tan^{-1} 0 = 0

Concept Problem Solution

The first step in this problem is to ask yourself "What angle would produce a cotangent of \sqrt{2} ?"

Since values for "x" and "y" around the unit circle are all fractions, and cotangent is equal to \frac{x}{y} , you need to find a pair of equations on the unit circle which, when divided by each other, give \sqrt{2} as the answer.

When looking around the unit circle, you can see that \cot 30^\circ = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}

Therefore, \cot^{-1} \sqrt{3} = 30^\circ

Then you can apply the next function:

\csc 30^\circ = \frac{hypotenuse}{opposite} = \frac{1}{\frac{1}{2}} = 2

And so

\csc (\cot^{-1} \sqrt{3}) = 2

Practice

Without using technology, find the exact value of each of the following. Use the restricted domain for each function.

  1. \sin \left ( \sec^{-1} \sqrt{2} \right )
  2. \cos \left ( \csc^{-1} 1 \right )
  3. \tan \left ( \cot^{-1} \sqrt{3} \right )
  4. \cos \left ( \csc^{-1} 2 \right )
  5. \cot \left ( \cos^{-1} 1 \right )
  6. \csc \left ( \sin^{-1} \frac{\sqrt{2}}{2} \right )
  7. \sec^{-1} \left ( \cos \pi \right )
  8. \cot^{-1} \left ( \tan \frac{\pi}{4} \right )
  9. \sec^{-1} \left ( \csc \frac{\pi}{4} \right )
  10. \csc^{-1} \left ( \sec \frac{\pi}{3} \right )
  11. \cos^{-1} \left ( \cot -\frac{\pi}{4} \right )
  12. \tan \left ( \cot^{-1} 0 \right )
  13. \sin \left ( \csc^{-1} \frac{2\sqrt{3}}{3} \right )
  14. \cot^{-1} \left ( \sin \frac{\pi}{2} \right )
  15. \cos \left ( \sec^{-1} \frac{2\sqrt{3}}{3} \right )

Image Attributions

Explore More

Sign in to explore more, including practice questions and solutions for Composition of Inverse Reciprocal Trig Functions.

Reviews

Please wait...
Please wait...

Original text