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# 4.3: Inverse Trigonometric Properties

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Relate the concept of inverse functions to trigonometric functions.
• Reduce the composite function to an algebraic expression involving no trigonometric functions.
• Use the inverse reciprocal properties.
• Compose each of the six basic trigonometric functions with $\tan^{-1} x$.

## Composing Trig Functions and their Inverses

In the Prerequisite Chapter, you learned that for a function $f(f^{-1}(x)) = x$ for all values of $x$ for which $f^{-1}(x)$ is defined. If this property is applied to the trigonometric functions, the following equations will be true whenever they are defined:

$\sin(\sin^{-1}(x)) = x && \cos(\cos^{-1}(x)) = x && \tan(\tan^{-1}(x)) = x$

As well, you learned that $f^{-1}(f(x)) = x$ for all values of $x$ for which $f(x)$ is defined. If this property is applied to the trigonometric functions, the following equations that deal with finding an inverse trig. function of a trig. function, will only be true for values of $x$ within the restricted domains.

$\sin^{-1}(\sin(x)) = x && \cos^{-1}(\cos(x)) = x && \tan^{-1}(\tan(x)) = x$

These equations are better known as composite functions and are composed of one trigonometric function in conjunction with another different trigonometric function. The composite functions will become algebraic functions and will not display any trigonometry. Let’s investigate this phenomenon.

Example 1: Find $\sin \left ( \sin^{-1} \frac{\sqrt{2}}{2} \right )$.

Solution: We know that $\sin^{-1} \frac{\sqrt{2}}{2} = \frac{\pi}{4}$, within the defined restricted domain. Then, we need to find $\sin \frac{\pi}{4}$, which is $\frac{\sqrt{2}}{2}$. So, the above properties allow for a short cut. $\sin \left ( \sin^{-1} \frac{\sqrt{2}}{2} \right ) = \frac{\sqrt{2}}{2}$, think of it like the sine and sine inverse cancel each other out and all that is left is the $\frac{\sqrt{2}}{2}$.

## Composing Trigonometric Functions

Besides composing trig functions with their own inverses, you can also compose any trig functions with any inverse. When solving these types of problems, start with the function that is composed inside of the other and work your way out. Use the following examples as a guideline.

Example 2: Without using technology, find the exact value of each of the following:

a. $\cos \left ( \tan^{-1} \sqrt{3} \right )$

b. $\tan \left ( \sin^{-1}\left( -\frac{1}{2} \right ) \right )$

c. $\cos (\tan^{-1} (-1))$

d. $\sin \left ( \cos^{-1}\frac{\sqrt{2}}{2} \right )$

Solution: For all of these types of problems, the answer is restricted to the inverse functions’ ranges.

a. $\cos \left ( \tan^{-1} \sqrt{3} \right )$: First find $\tan^{-1} \sqrt{3}$, which is $\frac{\pi}{3}$. Then find $\cos \frac{\pi}{3}$. Your final answer is $\frac{1}{2}$. Therefore, $\cos \left ( \tan^{-1} \sqrt{3} \right ) = \frac{1}{2}$.

b. $\tan \left ( \sin^{-1} \left ( -\frac{1}{2} \right ) \right ) = \tan \left ( -\frac{\pi}{6} \right ) = -\frac{\sqrt{3}}{3}$

c. $\cos (\tan^{-1} (-1)) = \cos^{-1} \left ( -\frac{\pi}{4} \right ) = \frac{\sqrt{2}}{2}$.

d. $\sin \left ( \cos^{-1} \frac{\sqrt{2}}{2} \right ) = \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}$

## Inverse Reciprocal Functions

We already know that the cosecant function is the reciprocal of the sine function. This will be used to derive the reciprocal of the inverse sine function.

$y & = \sin^{-1} x\\x & = \sin y\\\frac{1}{x} & = \csc y\\\csc^{-1} \frac{1}{x} & = y\\\csc^{-1} \frac{1}{x} & = \sin^{-1} x$

Because cosecant and secant are inverses, $\sin^{-1} \frac{1}{x} = \csc^{-1} x$ is also true.

The inverse reciprocal identity for cosine and secant can be proven by using the same process as above. However, remember that these inverse functions are defined by using restricted domains and the reciprocals of these inverses must be defined with the intervals of domain and range on which the definitions are valid.

$\sec^{-1} \frac{1}{x} = \cos^{-1} x \leftrightarrow \cos^{-1} \frac{1}{x} = \sec^{-1} x$

Tangent and cotangent have a slightly different relationship. Recall that the graph of cotangent differs from tangent by a reflection over the $y-$axis and a shift of $\frac{\pi}{2}$. As an equation, this can be written as $\cot x = -\tan \left ( x-\frac{\pi}{2} \right )$. Taking the inverse of this function will show the inverse reciprocal relationship between arccotangent and arctangent.

$y & = \cot^{-1} x\\y & = -\tan^{-1} \left ( x-\frac{\pi}{2} \right )\\x & = -\tan \left( y-\frac{\pi}{2} \right )\\-x & = \tan \left ( y-\frac{\pi}{2} \right )\\\tan^{-1} (-x) & = y-\frac{\pi}{2}\\\frac{\pi}{2} + \tan^{-1} (-x) & = y\\\frac{\pi}{2} - \tan^{-1} x & = y$

Remember that tangent is an odd function, so that $\tan(-x) = -\tan(x)$. Because tangent is odd, its inverse is also odd. So, this tells us that $\cot^{-1} x = \frac{\pi}{2} - \tan^{-1} x$ and $\tan^{-1} x =\frac{\pi}{2} - \cot^{-1} x$. You will determine the domain and range of all of these functions when you graph them in the exercises for this section. To graph arcsecant, arccosecant, and arccotangent in your calculator you will use these conversion identities: $\sec^{-1} x = \cos^{-1} \frac{1}{x} , \csc^{-1} x = \sin^{-1} \frac{1}{x} , \cot^{-1} x = \frac{\pi}{2} -\tan^{-1} x$. Note: It is also true that $\cot^{-1} x = \tan^{-1} \frac{1}{x}$.

Now, let’s apply these identities to some problems that will give us an insight into how they work.

Example 3: Evaluate $\sec^{-1}\sqrt{2}$

Solution: Use the inverse reciprocal property. $\sec^{-1} x = \cos^{-1} \frac{1}{x} \rightarrow \sec^{-1} \sqrt{2} = \cos^{-1} \frac{1}{\sqrt{2}}$. Recall that $\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$. So, $\sec^{-1} \sqrt{2} = \cos^{-1} \frac{\sqrt{2}}{2}$, and we know that $\cos^{-1} \frac{\sqrt{2}}{2} = \frac{\pi}{4}$. Therefore, $\sec^{-1} \sqrt{2} = \frac{\pi}{4}$.

Example 4: Find the exact value of each expression within the restricted domain, without a calculator.

a. $\sec^{-1} \sqrt{2}$

b. $\cot^{-1} \left ( -\sqrt{3} \right )$

c. $\csc^{-1} \frac{2\sqrt{3}}{3}$

Solution: For each of these problems, first find the reciprocal and then determine the angle from that.

a. $\sec^{-1} \sqrt{2} = \cos^{-1} \frac{\sqrt{2}}{2}$ From the unit circle, we know that the answer is $\frac{\pi}{4}$.

b. $\cot^{-1} \left ( -\sqrt{3} \right ) = \frac{\pi}{2} - \tan^{-1} \left ( -\sqrt{3} \right )$ From the unit circle, the answer is $\frac{5\pi}{6}$.

c. $\csc^{-1} \frac{2\sqrt{3}}{3} = \sin^{-1} \frac{\sqrt{3}}{2}$ Within our interval, there are is one answer, $\frac{\pi}{3}$.

Example 5: Using technology, find the value in radian measure, of each of the following:

a. $\arcsin 0.6384$

b. $\arccos (-0.8126)$

c. $\arctan (-1.9249)$

Solution:

a.

b.

c.

## Composing Inverse Reciprocal Trig Functions

In this subsection, we will combine what was learned in the previous two sections. Here are a few examples:

Example 6: 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 7: 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 8: 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 $\cos \theta = \frac{4}{5}$.

Example 9: Evaluate $\tan \left ( \sin^{-1} \left ( -\frac{3}{4} \right ) \right )$

Solution: Even though $\frac{3}{4}$ does not represent two lengths from a Pythagorean Triple, you can still use the Pythagorean Theorem to find the missing side. $(-3)^2 + b^2 = 4^2$, so $b = \sqrt{16-9} = \sqrt{7}$. From the restricted domain, sine inverse is negative in the $4^{th}$ Quadrant. To illustrate:

Let $\theta & = \sin^{-1} \left ( -\frac{3}{4} \right )\\\sin \theta & = -\frac{3}{4}\\\tan \left ( \sin^{-1} \left ( -\frac{3}{4} \right ) \right ) & = \tan \theta\\\tan \theta & = \frac{-3}{\sqrt{7}} \ \text{or} \ \frac{-3\sqrt{7}}{7}$

## Trigonometry in Terms of Algebra

All of the trigonometric functions can be rewritten in terms of only $x$, when using one of the inverse trigonometric functions. Starting with tangent, we draw a triangle where the opposite side (from $\theta$) is defined as $x$ and the adjacent side is 1. The hypotenuse, from the Pythagorean Theorem would be $\sqrt{x^2+1}$. Substituting $\tan^{-1} x$ for $\theta$, we get:

$\tan \theta & = \frac{x}{1}\\\tan \theta & = x && hypotenuse = \sqrt{x^2+1}\\\theta & = \tan^{-1} x$

$\sin (\tan^{-1}x) & = \sin \theta = \frac{x}{\sqrt{x^2+1}} && \csc (\tan^{-1}x) = \csc \theta = \frac{\sqrt{x^2+1}}{x}\\\cos (\tan^{-1}x) & = \cos \theta = \frac{1}{\sqrt{x^2+1}} && \sec (\tan^{-1}x) = \sec \theta = \sqrt{x^2+1}\\\tan (\tan^{-1}x) & = \tan \theta = x && \cot (\tan^{-1}x) = \cot \theta = \frac{1}{x}$

Example 10: Find $\sin (\tan^{-1} 3x)$.

Solution: Instead of using $x$ in the ratios above, use $3x$.

$\sin (\tan^{-1} 3x) = \sin \theta = \frac{3x}{\sqrt{(3x)^2 + 1}} = \frac{3x}{\sqrt{9x^2+1}}$

Example 11: Find $\sec^2 (\tan^{-1} x)$.

Solution: This problem might be better written as $[\sec (\tan^{-1} x)]^2$. Therefore, all you need to do is square the ratio above.

$[\sec (\tan^{-1}x)]^2 = \left ( \sqrt{x^2+1} \right )^2 = x^2 + 1$

You can also write the all of the trig functions in terms of arcsine and arccosine. However, for each inverse function, there is a different triangle. You will derive these formulas in the exercise for this section.

## Points to Consider

• Is it possible to graph these composite functions? What happens when you graph $y = \sin (\cos^{-1} x)$ in your calculator?
• Do exact values of functions of inverse functions exist if any value is used?

## Review Questions

1. Find the exact value of the functions, without a calculator, over their restricted domains.
1. $\cos^{-1} \frac{\sqrt{3}}{2}$
2. $\sec^{-1} \sqrt{2}$
3. $\sec^{-1} \left ( -\sqrt{2} \right )$
4. $\sec^{-1} (-2)$
5. $\cot^{-1} (-1)$
6. $\csc^{-1} \left ( \sqrt{2} \right )$
2. Use your calculator to find:
1. $\arccos (-0.923)$
2. $\arcsin 0.368$
3. $\arctan 5.698$
3. Find the exact value of the functions, without a calculator, over their restricted domains.
1. $\csc \left ( \cos^{-1} \frac{\sqrt{3}}{2} \right )$
2. $\sec^{-1} ( \tan (\cot^{-1} 1))$
3. $\tan^{-1} \left ( \cos \frac{\pi}{2} \right )$
4. $\cot \left ( \sec^{-1} \frac{2\sqrt{3}}{3} \right )$
4. Using your graphing calculator, graph $y = \sec^{-1} x$. Sketch this graph, determine the domain and range, $x-$ and/or $y-$intercepts. (Your calculator knows the restriction on this function, there is no need to input it into $Y =$.)
5. Using your graphing calculator, graph $y = \csc^{-1} x$. Sketch this graph, determine the domain and range, $x-$ and/or $y-$intercepts. (Your calculator knows the restriction on this function, there is no need to input it into $Y =$.)
6. Using your graphing calculator, graph $y = \cot^{-1} x$. Sketch this graph, determine the domain and range, $x-$ and/or $y-$intercepts. (Your calculator knows the restriction on this function, there is no need to input it into $Y =$.)
7. Evaluate:
1. $\sin \left ( \cos^{-1} \frac{5}{13} \right )$
2. $\tan \left ( \sin^{-1} \left( -\frac{6}{11} \right) \right )$
3. $\cos \left ( \csc^{-1} \frac{25}{7} \right )$
8. Express each of the following functions as an algebraic expression involving no trigonometric functions.
1. $\cos^2 (\tan^{-1} x)$
2. $\cot (\tan^{-1} x^2)$
9. To find trigonometric functions in terms of sine inverse, use the following triangle.
1. Determine the sine, cosine and tangent in terms of arcsine.
2. Find $\tan (\sin^{-1} 2x^3)$.
10. To find the trigonometric functions in terms of cosine inverse, use the following triangle.
1. Determine the sine, cosine and tangent in terms of arccosine.
2. Find $\sin^2 \left ( \cos^{-1} \frac{1}{2}x \right )$.

1. $\frac{\pi}{6}$
2. $\frac{\pi}{4}$
3. $\frac{3\pi}{4}$
4. $\frac{2\pi}{3}$
5. $-\frac{\pi}{4}$
6. $\frac{\pi}{4}$
1. 2.747
2. 0.377
3. 1.397
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. $\tan^{-1} \left ( \cos \frac{\pi}{2} \right ) = \tan^{-1} 0 = 0$
4. $\cot \left ( \sec^{-1} \frac{2\sqrt{3}}{3} \right ) = \cot \left ( \cos^{-1} \frac{\sqrt{3}}{2} \right ) = \cot \frac{\pi}{6} = \frac{1}{\tan \frac{\pi}{6}} = \frac{1}{\frac{\sqrt{3}}{3}} = \sqrt{3}$
1. $y = \sec^{-1} x$ when plugged into your graphing calculator is $y = \cos^{-1} \frac{1}{x}$. The domain is all reals, excluding the interval (-1, 1). The range is all reals in the interval $[0, \pi], y \ne \frac{\pi}{2}$. There are no $y$ intercepts and the only $x$ intercept is at 1.
2. $y = \csc^{-1} x$ when plugged into your graphing calculator is $y = \sin^{-1} \frac{1}{x}$. The domain is all reals, excluding the interval (-1, 1). The range is all reals in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}], y \ne 0$. There are no $x$ or $y$ intercepts.
3. The domain is all real numbers and the range is from $(0, \pi)$. There is an $x-$intercept at $\frac{\pi}{2}$.
1. $\cos \theta & = \frac{5}{13}\\\sin \left ( \cos^{-1} \left ( \frac{5}{13} \right ) \right ) & = \sin \theta\\\sin \theta & = \frac{12}{13}$
2. $\tan \left ( \sin^{-1} \left ( -\frac{6}{11} \right ) \right ) \rightarrow \sin \theta = -\frac{6}{11}$. The third side is $b = \sqrt{121-36} = \sqrt{85}$. $\tan \theta = -\frac{6}{\sqrt{85}} = -\frac{6\sqrt{85}}{85}$
3. $\cos \left ( \csc^{-1} \left ( \frac{25}{7} \right ) \right ) \rightarrow \csc \theta = \frac{25}{7} \rightarrow \sin \theta = \frac{7}{25}$. This two lengths of a Pythagorean Triple, with the third side being 24. $\cos \theta = \frac{24}{25}$
1. $\frac{1}{x^2+1}$
2. $\frac{1}{x^2}$
4. The adjacent side to $\theta$ is $\sqrt{1-x^2}$, so the three trig functions are:
1. $\sin (\sin^{-1} x) & = \sin \theta = x\\\cos (\sin^{-1}x) & = \cos \theta = \sqrt{1-x^2}\\\tan (\sin^{-1}x) & = \tan \theta = \frac{x}{\sqrt{1-x^2}}$
2. $\tan(\sin^{-1} (2x^3)) = \frac{2x^3}{\sqrt{1-(2x^3)^2}} = \frac{2x^3}{\sqrt{1-4x^6}}$
5. The opposite side to $\theta$ is $\sqrt{1-x^2}$, so the three trig functions are:
1. $\sin (\cos^{-1} x) & = \sin \theta = \sqrt{1-x^2}\\\cos (\cos^{-1}x) & = \cos \theta = x\\\tan (\cos^{-1}x) & = \tan \theta = \frac{\sqrt{1-x^2}}{x}$
2. $\sin^2 \left ( \cos^{-1} \left ( \frac{1}{2}x \right ) \right ) = \left ( \sqrt{1-\left ( \frac{1}{2}x \right )^2} \right )^2 = 1-\frac{1}{4}x^2$

Feb 23, 2012

Apr 23, 2015