Understanding and Using the Inverse Sine, Cosine, and Tangent Functions

In order to use inverse trigonometric functions, we need to understand that an inverse trigonometric function “undoes” what the original trigonometric function “does,” as is the case with any other function and its inverse. In other words, the domain of the inverse function is the range of the original function, and vice versa, as summarized in (Figure).

Figure 1.

For example, if$\,f\left(x\right)=\mathrm{sin}\,x,\,$then we would write$\,{f}^{-1}\left(x\right)={\mathrm{sin}}^{-1}x.\,$Be aware that$\,{\mathrm{sin}}^{-1}x\,$does not mean$\,\frac{1}{\mathrm{sin}x}.\,$The following examples illustrate the inverse trigonometric functions:

• Since$\,\text{sin}\left(\frac{\pi }{6}\right)=\frac{1}{2},\,$then$\,\frac{\pi }{6}={\text{sin}}^{-1}\left(\frac{1}{2}\right).$
• Since$\,\mathrm{cos}\left(\pi \right)=-1,\,$then$\,\pi ={\mathrm{cos}}^{-1}\left(-1\right).$
• Since$\,\mathrm{tan}\left(\frac{\pi }{4}\right)=1,\,$then$\,\frac{\pi }{4}={\mathrm{tan}}^{-1}\left(1\right).$

In previous sections, we evaluated the trigonometric functions at various angles, but at times we need to know what angle would yield a specific sine, cosine, or tangent value. For this, we need inverse functions. Recall that, for a one-to-one function, if$\,f\left(a\right)=b,\,$then an inverse function would satisfy$\,{f}^{-1}\left(b\right)=a.$

Bear in mind that the sine, cosine, and tangent functions are not one-to-one functions. The graph of each function would fail the horizontal line test. In fact, no periodic function can be one-to-one because each output in its range corresponds to at least one input in every period, and there are an infinite number of periods. As with other functions that are not one-to-one, we will need to restrict the domain of each function to yield a new function that is one-to-one. We choose a domain for each function that includes the number 0. (Figure) shows the graph of the sine function limited to$\,\left[-\frac{\pi }{2},\frac{\pi }{2}\right]\,$and the graph of the cosine function limited to$\,\left[0,\pi \right].$

Figure 2. (a) Sine function on a restricted domain of$\,\left[-\frac{\pi }{2},\frac{\pi }{2}\right];\,$(b) Cosine function on a restricted domain of$\,\left[0,\pi \right]$

(Figure) shows the graph of the tangent function limited to$\,\left(-\frac{\pi }{2},\frac{\pi }{2}\right).$

Figure 3. Tangent function on a restricted domain of$\,\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$

These conventional choices for the restricted domain are somewhat arbitrary, but they have important, helpful characteristics. Each domain includes the origin and some positive values, and most importantly, each results in a one-to-one function that is invertible. The conventional choice for the restricted domain of the tangent function also has the useful property that it extends from one vertical asymptote to the next instead of being divided into two parts by an asymptote.

On these restricted domains, we can define the inverse trigonometric functions.

• The inverse sine function$\,y={\mathrm{sin}}^{-1}x\,$means$\,x=\mathrm{sin}\,y.\,$The inverse sine function is sometimes called the arcsine function, and notated$\,\mathrm{arcsin}x.$
  $y={\mathrm{sin}}^{-1}x\,\text{has domain}\,\left[-1,1\right]\,\text{and range}\,\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$
• The inverse cosine function$\,y={\mathrm{cos}}^{-1}x\,$means$\,x=\mathrm{cos}\,y.\,$The inverse cosine function is sometimes called the arccosine function, and notated$\,\mathrm{arccos}\,x.$
  $y={\mathrm{cos}}^{-1}x\,\text{has domain}\,\left[-1,1\right]\,\text{and range}\,\left[0,\pi \right]$
• The inverse tangent function$\,y={\mathrm{tan}}^{-1}x\,$means$\,x=\mathrm{tan}\,y.\,$The inverse tangent function is sometimes called the arctangent function, and notated$\,\mathrm{arctan}\,x.$
  $y={\mathrm{tan}}^{-1}x\,\text{has domain}\,\left(\mathrm{-\infty },\infty \right)\,\text{and range}\,\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$

The graphs of the inverse functions are shown in (Figure), (Figure), and (Figure). Notice that the output of each of these inverse functions is a number, an angle in radian measure. We see that$\,{\mathrm{sin}}^{-1}x\,$has domain$\,\left[-1,1\right]\,$and range$\,\left[-\frac{\pi }{2},\frac{\pi }{2}\right],$${\mathrm{cos}}^{-1}x\,$has domain$\,\left[-1,1\right]\,$and range$\,\left[0,\pi \right],\,$and$\,{\mathrm{tan}}^{-1}x\,$has domain of all real numbers and range$\,\left(-\frac{\pi }{2},\frac{\pi }{2}\right).\,$To find the domain and range of inverse trigonometric functions, switch the domain and range of the original functions. Each graph of the inverse trigonometric function is a reflection of the graph of the original function about the line$\,y=x.$

Figure 4. The sine function and inverse sine (or arcsine) function

Figure 5. The cosine function and inverse cosine (or arccosine) function

Figure 6. The tangent function and inverse tangent (or arctangent) function

Finding the Exact Value of Expressions Involving the Inverse Sine, Cosine, and Tangent Functions

Now that we can identify inverse functions, we will learn to evaluate them. For most values in their domains, we must evaluate the inverse trigonometric functions by using a calculator, interpolating from a table, or using some other numerical technique. Just as we did with the original trigonometric functions, we can give exact values for the inverse functions when we are using the special angles, specifically$\,\frac{\pi }{6}\,$(30°),$\,\frac{\pi }{4}\,$(45°), and$\,\frac{\pi }{3}\,$(60°), and their reflections into other quadrants.

Using a Calculator to Evaluate Inverse Trigonometric Functions

To evaluate inverse trigonometric functions that do not involve the special angles discussed previously, we will need to use a calculator or other type of technology. Most scientific calculators and calculator-emulating applications have specific keys or buttons for the inverse sine, cosine, and tangent functions. These may be labeled, for example, SIN
${\text{}}^{-1}$, ARCSIN, or ASIN.

In the previous chapter, we worked with trigonometry on a right triangle to solve for the sides of a triangle given one side and an additional angle. Using the inverse trigonometric functions, we can solve for the angles of a right triangle given two sides, and we can use a calculator to find the values to several decimal places.

In these examples and exercises, the answers will be interpreted as angles and we will use$\,\theta \,$as the independent variable. The value displayed on the calculator may be in degrees or radians, so be sure to set the mode appropriate to the application.

How To

Given two sides of a right triangle like the one shown in (Figure), find an angle.

Figure 7.

1. If one given side is the hypotenuse of length$\,h\,$and the side of length$\,a\,$adjacent to the desired angle is given, use the equation$\,\,\theta ={\mathrm{cos}}^{-1}\left(\frac{a}{h}\right).$
2. If one given side is the hypotenuse of length$\,h\,$and the side of length$\,p\,$opposite to the desired angle is given, use the equation$\,\theta ={\mathrm{sin}}^{-1}\left(\frac{p}{h}\right).$
3. If the two legs (the sides adjacent to the right angle) are given, then use the equation$\,\theta ={\mathrm{tan}}^{-1}\left(\frac{p}{a}\right).$

Applying the Inverse Cosine to a Right Triangle

Solve the triangle in (Figure) for the angle$\,\theta .$

Figure 8.

Show Solution

Because we know the hypotenuse and the side adjacent to the angle, it makes sense for us to use the cosine function.

$\begin{array}{ll}\mathrm{cos}\,\theta =\frac{9}{12}\hfill & \begin{array}{ccc}& & \end{array}\hfill \\ \,\,\,\,\,\,\,\,\text{ }\theta ={\mathrm{cos}}^{-1}\left(\frac{9}{12}\right)\hfill & \begin{array}{ccc}& & \end{array}\text{Apply definition of the inverse}.\hfill \\ \,\,\,\,\,\,\,\,\text{ }\theta \approx 0.7227\text{ or about }41.4096°\hfill & \begin{array}{ccc}& & \end{array}\text{Evaluate}.\hfill \end{array}$

Try It

Solve the triangle in (Figure) for the angle$\,\theta .$

Figure 9.

Show Solution

${\mathrm{sin}}^{-1}\left(0.6\right)=36.87°=0.6435\,$radians

Finding Exact Values of Composite Functions with Inverse Trigonometric Functions

There are times when we need to compose a trigonometric function with an inverse trigonometric function. In these cases, we can usually find exact values for the resulting expressions without resorting to a calculator. Even when the input to the composite function is a variable or an expression, we can often find an expression for the output. To help sort out different cases, let$\,f\left(x\right)\,$and$\,g\left(x\right)\,$be two different trigonometric functions belonging to the set$\,\left\{\mathrm{sin}\left(x\right),\mathrm{cos}\left(x\right),\mathrm{tan}\left(x\right)\right\}\,$and let$\,{f}^{-1}\left(y\right)\,$and$\,{g}^{-1}\left(y\right)$be their inverses.

Evaluating Compositions of the Form f(f⁻¹(y)) and f⁻¹(f(x))

For any trigonometric function,$\,f\left({f}^{-1}\left(y\right)\right)=y\,$for all$\,y\,$in the proper domain for the given function. This follows from the definition of the inverse and from the fact that the range of$\,f\,$was defined to be identical to the domain of$\,{f}^{-1}.\,$However, we have to be a little more careful with expressions of the form$\,{f}^{-1}\left(f\left(x\right)\right).$

Compositions of a trigonometric function and its inverse

[latex]\begin{array}{l}\,\,\mathrm{sin}\left({\mathrm{sin}}^{-1}x\right)=x\,\text{for}\,-1\le x\le 1\hfill \\ \mathrm{cos}\left({\mathrm{cos}}^{-1}x\right)=x\,\text{for}\,-1\le x\le 1\hfill \\ \,\mathrm{tan}\left({\mathrm{tan}}^{-1}x\right)=x\,\text{for}\,-\infty

[latex]\begin{array}{l}\,{\mathrm{sin}}^{-1}\left(\mathrm{sin}\,x\right)=x\,\text{only for }-\frac{\pi }{2}\le x\le \frac{\pi }{2}\hfill \\ {\mathrm{cos}}^{-1}\left(\mathrm{cos}\,x\right)=x\,\text{only for }0\le x\le \pi \hfill \\ \,{\mathrm{tan}}^{-1}\left(\mathrm{tan}\,x\,\right)=x\,\text{only for }-\frac{\pi }{2}

Is it correct that$\,{\mathrm{sin}}^{-1}\left(\mathrm{sin}\,x\right)=x?$

No. This equation is correct if$\,x\,$belongs to the restricted domain$\,\left[-\frac{\pi }{2},\frac{\pi }{2}\right],\,$but sine is defined for all real input values, and for$\,x\,$outside the restricted interval, the equation is not correct because its inverse always returns a value in$\,\left[-\frac{\pi }{2},\frac{\pi }{2}\right].\,$The situation is similar for cosine and tangent and their inverses. For example,$\,{\mathrm{sin}}^{-1}\left(\mathrm{sin}\left(\frac{3\pi }{4}\right)\right)=\frac{\pi }{4}.$

Given an expression of the form f⁻¹(f(θ)) where$\,f\left(\theta \right)=\mathrm{sin}\,\theta ,\text{ }\mathrm{cos}\,\theta ,\text{ or }\mathrm{tan}\,\theta ,\,$evaluate.

1. If$\,\theta \,$is in the restricted domain of$\,f,\text{ then }{f}^{-1}\left(f\left(\theta \right)\right)=\theta .$
2. If not, then find an angle$\,\varphi \,$within the restricted domain of$\,f\,$such that$\,f\left(\varphi \right)=f\left(\theta \right).\,$Then$\,{f}^{-1}\left(f\left(\theta \right)\right)=\varphi .$

Using Inverse Trigonometric Functions

Evaluate the following:

1. ${\mathrm{sin}}^{-1}\left(\mathrm{sin}\left(\frac{\pi }{3}\right)\right)$
2. ${\mathrm{sin}}^{-1}\left(\mathrm{sin}\left(\frac{2\pi }{3}\right)\right)$
3. ${\mathrm{cos}}^{-1}\left(\mathrm{cos}\left(\frac{2\pi }{3}\right)\right)$
4. ${\mathrm{cos}}^{-1}\left(\mathrm{cos}\left(-\frac{\pi }{3}\right)\right)$

Show Solution

1. $\frac{\pi }{3}\text{ is in }\left[-\frac{\pi }{2},\frac{\pi }{2}\right],\,$so$\,{\mathrm{sin}}^{-1}\left(\mathrm{sin}\left(\frac{\pi }{3}\right)\right)=\frac{\pi }{3}.$
2. $\frac{2\pi }{3}\text{ is not in }\left[-\frac{\pi }{2},\frac{\pi }{2}\right],\,$but$\,\mathrm{sin}\left(\frac{2\pi }{3}\right)=\mathrm{sin}\left(\frac{\pi }{3}\right),\,$so$\,{\mathrm{sin}}^{-1}\left(\mathrm{sin}\left(\frac{2\pi }{3}\right)\right)=\frac{\pi }{3}.$
3. $\frac{2\pi }{3}\text{ is in }\left[0,\pi \right],\,$so$\,{\mathrm{cos}}^{-1}\left(\mathrm{cos}\left(\frac{2\pi }{3}\right)\right)=\frac{2\pi }{3}.$
4. $-\frac{\pi }{3}\text{ is not in }\left[0,\pi \right],\,$but$\,\mathrm{cos}\left(-\frac{\pi }{3}\right)=\mathrm{cos}\left(\frac{\pi }{3}\right)\,$because cosine is an even function.
5. $\frac{\pi }{3}\text{ is in }\left[0,\pi \right],\,$so$\,{\mathrm{cos}}^{-1}\left(\mathrm{cos}\left(-\frac{\pi }{3}\right)\right)=\frac{\pi }{3}.$

Try It

Evaluate$\,{\mathrm{tan}}^{-1}\left(\mathrm{tan}\left(\frac{\pi }{8}\right)\right)\,\text{and}\,{\mathrm{tan}}^{-1}\left(\mathrm{tan}\left(\frac{11\pi }{9}\right)\right).$

Show Solution

$\frac{\pi }{8};\frac{2\pi }{9}$

Evaluating Compositions of the Form f⁻¹(g(x))

Now that we can compose a trigonometric function with its inverse, we can explore how to evaluate a composition of a trigonometric function and the inverse of another trigonometric function. We will begin with compositions of the form$\,{f}^{-1}\left(g\left(x\right)\right).\,$For special values of$\,x,$we can exactly evaluate the inner function and then the outer, inverse function. However, we can find a more general approach by considering the relation between the two acute angles of a right triangle where one is$\,\theta ,\,$making the other$\,\frac{\pi }{2}-\theta .$Consider the sine and cosine of each angle of the right triangle in (Figure).

Figure 10. Right triangle illustrating the cofunction relationships

Because$\,\mathrm{cos}\,\theta =\frac{b}{c}=\mathrm{sin}\left(\frac{\pi }{2}-\theta \right),\,$we have$\,{\mathrm{sin}}^{-1}\left(\mathrm{cos}\,\theta \right)=\frac{\pi }{2}-\theta \,$if$\,0\le \theta \le \pi .\,$If$\,\theta \,$is not in this domain, then we need to find another angle that has the same cosine as$\,\theta \,$and does belong to the restricted domain; we then subtract this angle from$\,\frac{\pi }{2}.$Similarly,$\,\mathrm{sin}\,\theta =\frac{a}{c}=\mathrm{cos}\left(\frac{\pi }{2}-\theta \right),\,$so$\,{\mathrm{cos}}^{-1}\left(\mathrm{sin}\,\theta \right)=\frac{\pi }{2}-\theta \,$if$\,-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}.\,$These are just the function-cofunction relationships presented in another way.

Given functions of the form$\,{\mathrm{sin}}^{-1}\left(\mathrm{cos}\,x\right)\,$and$\,{\mathrm{cos}}^{-1}\left(\mathrm{sin}\,x\right),\,$evaluate them.

1. If$\,x\text{ is in }\left[0,\pi \right],\,$then$\,{\mathrm{sin}}^{-1}\left(\mathrm{cos}\,x\right)=\frac{\pi }{2}-x.$
2. If$\,x\text{ is not in }\left[0,\pi \right],\,$then find another angle$\,y\text{ in }\left[0,\pi \right]\,$such that$\,\mathrm{cos}\,y=\mathrm{cos}\,x.$
   ${\mathrm{sin}}^{-1}\left(\mathrm{cos}\,x\right)=\frac{\pi }{2}-y$
3. If$\,x\text{ is in }\left[-\frac{\pi }{2},\frac{\pi }{2}\right],\,$then$\,{\mathrm{cos}}^{-1}\left(\mathrm{sin}\,x\right)=\frac{\pi }{2}-x.$
4. If$\,x\text{ is not in}\,\left[-\frac{\pi }{2},\frac{\pi }{2}\right],\,$then find another angle$\,y\text{ in }\left[-\frac{\pi }{2},\frac{\pi }{2}\right]\,$such that$\,\mathrm{sin}\,y=\mathrm{sin}\,x.$
   ${\mathrm{cos}}^{-1}\left(\mathrm{sin}\,x\right)=\frac{\pi }{2}-y$

Evaluating the Composition of an Inverse Sine with a Cosine

Evaluate$\,{\mathrm{sin}}^{-1}\left(\mathrm{cos}\left(\frac{13\pi }{6}\right)\right)$

1. by direct evaluation.
2. by the method described previously.

Show Solution

1. Here, we can directly evaluate the inside of the composition.
   $\begin{array}{l}\hfill \\ \begin{array}{l}\mathrm{cos}\left(\frac{13\pi }{6}\right)=\mathrm{cos}\left(\frac{\pi }{6}+2\pi \right)\hfill \\ \text{ }=\mathrm{cos}\left(\frac{\pi }{6}\right)\hfill \\ \text{ }=\frac{\sqrt{3}}{2}\hfill \end{array}\hfill \end{array}$

   Now, we can evaluate the inverse function as we did earlier.

   ${\mathrm{sin}}^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi }{3}$
2. We have$\,x=\frac{13\pi }{6}\text{,}\,y=\frac{\pi }{6},\,$and
   $\begin{array}{r}\hfill {\mathrm{sin}}^{-1}\left(\mathrm{cos}\left(\frac{13\pi }{6}\right)\right)=\frac{\pi }{2}-\frac{\pi }{6}\\ \hfill =\frac{\pi }{3}\text{ }\end{array}$

Try It

Evaluate$\,{\mathrm{cos}}^{-1}\left(\mathrm{sin}\left(-\frac{11\pi }{4}\right)\right).$

Show Solution

$\frac{3\pi }{4}$

Evaluating Compositions of the Form f(g⁻¹(x))

To evaluate compositions of the form$\,f\left({g}^{-1}\left(x\right)\right),\,$where$\,f\,$and$\,g\,$are any two of the functions sine, cosine, or tangent and$\,x\,$is any input in the domain of$\,{g}^{-1},\,$we have exact formulas, such as$\,\mathrm{sin}\left({\mathrm{cos}}^{-1}x\right)=\sqrt{1-{x}^{2}}.\,$When we need to use them, we can derive these formulas by using the trigonometric relations between the angles and sides of a right triangle, together with the use of Pythagoras’s relation between the lengths of the sides. We can use the Pythagorean identity,$\,{\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x=1,\,$to solve for one when given the other. We can also use the inverse trigonometric functions to find compositions involving algebraic expressions.

Evaluating the Composition of a Sine with an Inverse Cosine

Find an exact value for$\,\mathrm{sin}\left({\mathrm{cos}}^{-1}\left(\frac{4}{5}\right)\right).$

Show Solution

Beginning with the inside, we can say there is some angle such that$\,\theta ={\mathrm{cos}}^{-1}\left(\frac{4}{5}\right),\,$which means$\,\mathrm{cos}\,\theta =\frac{4}{5},\,$and we are looking for$\,\mathrm{sin}\,\theta .\,$We can use the Pythagorean identity to do this.

$\begin{array}{llll}{\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta =1\hfill & \hfill & \hfill & \text{Use our known value for cosine}.\hfill \\ \,\,\,{\mathrm{sin}}^{2}\theta +{\left(\frac{4}{5}\right)}^{2}=1\hfill & \hfill & \hfill & \text{Solve for sine}.\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\mathrm{sin}}^{2}\theta =1-\frac{16}{25}\hfill & \hfill & \hfill & \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\mathrm{sin}\,\theta =±\sqrt{\frac{9}{25}}=±\frac{3}{5}\hfill & \hfill & \hfill & \hfill \end{array}$

Since$\,\theta ={\mathrm{cos}}^{-1}\left(\frac{4}{5}\right)\,$is in quadrant I,$\,\mathrm{sin}\,\theta \,$must be positive, so the solution is$\,\frac{3}{5}.\,$See (Figure).

Figure 11. Right triangle illustrating that if$\,\mathrm{cos}\,\theta =\frac{4}{5},\,$then$\,\mathrm{sin}\,\theta =\frac{3}{5}\,$

We know that the inverse cosine always gives an angle on the interval$\,\left[0,\pi \right],\,$so we know that the sine of that angle must be positive; therefore$\,\mathrm{sin}\left({\mathrm{cos}}^{-1}\left(\frac{4}{5}\right)\right)=\mathrm{sin}\,\theta =\frac{3}{5}.$

Try It

Evaluate$\,\mathrm{cos}\left({\mathrm{tan}}^{-1}\left(\frac{5}{12}\right)\right).$

Show Solution

$\frac{12}{13}$

Evaluating the Composition of a Sine with an Inverse Tangent

Find an exact value for$\,\mathrm{sin}\left({\mathrm{tan}}^{-1}\left(\frac{7}{4}\right)\right).$

Show Solution

While we could use a similar technique as in (Figure), we will demonstrate a different technique here. From the inside, we know there is an angle such that$\,\mathrm{tan}\,\theta =\frac{7}{4}.\,$We can envision this as the opposite and adjacent sides on a right triangle, as shown in (Figure).

Figure 12. A right triangle with two sides known

Using the Pythagorean Theorem, we can find the hypotenuse of this triangle.

$\begin{array}{l}\begin{array}{l}\hfill \\ \text{ }{4}^{2}+{7}^{2}={\text{hypotenuse}}^{2}\hfill \end{array}\hfill \\ \text{hypotenuse}=\sqrt{65}\hfill \end{array}$

Now, we can evaluate the sine of the angle as the opposite side divided by the hypotenuse.

$\mathrm{sin}\,\theta =\frac{7}{\sqrt{65}}$

This gives us our desired composition.

$\begin{array}{l}\mathrm{sin}\left({\mathrm{tan}}^{-1}\left(\frac{7}{4}\right)\right)=\mathrm{sin}\,\theta \hfill \\ \text{ }=\frac{7}{\sqrt{65}}\hfill \\ \text{ }=\frac{7\sqrt{65}}{65}\hfill \end{array}$

Try It

Evaluate$\,\mathrm{cos}\left({\mathrm{sin}}^{-1}\left(\frac{7}{9}\right)\right).$

Show Solution

$\frac{4\sqrt{2}}{9}$

Finding the Cosine of the Inverse Sine of an Algebraic Expression

Find a simplified expression for$\,\mathrm{cos}\left({\mathrm{sin}}^{-1}\left(\frac{x}{3}\right)\right)\,$for$\,-3\le x\le 3.$

Show Solution

We know there is an angle$\,\theta \,$such that$\,\mathrm{sin}\,\theta =\frac{x}{3}.$

$\begin{array}{ll}{\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta =1\hfill & \text{Use the Pythagorean Theorem}.\hfill \\ {\left(\frac{x}{3}\right)}^{2}+{\mathrm{cos}}^{2}\theta =1\hfill & \text{Solve for cosine}.\hfill \\ \text{ }\,{\mathrm{cos}}^{2}\theta =1-\frac{{x}^{2}}{9}\hfill & \hfill \\ \text{ }\,\text{ }\mathrm{cos}\theta =±\sqrt{\frac{9-{x}^{2}}{9}}=±\frac{\sqrt{9-{x}^{2}}}{3}\hfill & \hfill \end{array}$

Because we know that the inverse sine must give an angle on the interval$\,\left[-\frac{\pi }{2},\frac{\pi }{2}\right],\,$we can deduce that the cosine of that angle must be positive.

$\mathrm{cos}\left({\mathrm{sin}}^{-1}\left(\frac{x}{3}\right)\right)=\frac{\sqrt{9-{x}^{2}}}{3}$

Try It

Find a simplified expression for$\,\mathrm{sin}\left({\mathrm{tan}}^{-1}\left(4x\right)\right)\,$for$\,-\frac{1}{4}\le x\le \frac{1}{4}.$

Show Solution

$\frac{4x}{\sqrt{16{x}^{2}+1}}$

Access this online resource for additional instruction and practice with inverse trigonometric functions.

• Evaluate Expressions Involving Inverse Trigonometric Functions

Visit this website for additional practice questions from Learningpod.

Key Concepts

• An inverse function is one that “undoes” another function. The domain of an inverse function is the range of the original function and the range of an inverse function is the domain of the original function.
• Because the trigonometric functions are not one-to-one on their natural domains, inverse trigonometric functions are defined for restricted domains.
• For any trigonometric function$\,f\left(x\right),\,$if$\,x={f}^{-1}\left(y\right),\,$then$\,f\left(x\right)=y.\,$However,$\,f\left(x\right)=y\,$only implies$\,x={f}^{-1}\left(y\right)\,$if$\,x\,$is in the restricted domain of$\,f.\,$See (Figure).
• Special angles are the outputs of inverse trigonometric functions for special input values; for example,$\,\frac{\pi }{4}={\mathrm{tan}}^{-1}\left(1\right)\,\text{and}\,\frac{\pi }{6}={\mathrm{sin}}^{-1}\left(\frac{1}{2}\right).$See (Figure).
• A calculator will return an angle within the restricted domain of the original trigonometric function. See (Figure).
• Inverse functions allow us to find an angle when given two sides of a right triangle. See (Figure).
• In function composition, if the inside function is an inverse trigonometric function, then there are exact expressions; for example,$\,\mathrm{sin}\left({\mathrm{cos}}^{-1}\left(x\right)\right)=\sqrt{1-{x}^{2}}.\,$See (Figure).
• If the inside function is a trigonometric function, then the only possible combinations are$\,{\mathrm{sin}}^{-1}\left(\mathrm{cos}\,x\right)=\frac{\pi }{2}-x\,$if$\,0\le x\le \pi \,$and$\,{\mathrm{cos}}^{-1}\left(\mathrm{sin}\,x\right)=\frac{\pi }{2}-x\,$if$\,-\frac{\pi }{2}\le x\le \frac{\pi }{2}.$ See (Figure) and (Figure).
• When evaluating the composition of a trigonometric function with an inverse trigonometric function, draw a reference triangle to assist in determining the ratio of sides that represents the output of the trigonometric function. See (Figure).
• When evaluating the composition of a trigonometric function with an inverse trigonometric function, you may use trig identities to assist in determining the ratio of sides. See (Figure).

Section Exercises

Verbal

Why do the functions$\,f\left(x\right)={\mathrm{sin}}^{-1}x\,$and$\,g\left(x\right)={\mathrm{cos}}^{-1}x\,$have different ranges?

Show Solution

The function$\,y=\mathrm{sin}x\,$is one-to-one on$\,\left[-\frac{\pi }{2},\frac{\pi }{2}\right];\,$thus, this interval is the range of the inverse function of$\,y=\mathrm{sin}x,$$f\left(x\right)={\mathrm{sin}}^{-1}x.\,$The function$\,y=\mathrm{cos}x\,$is one-to-one on $\,\left[0,\pi \right];\,$thus, this interval is the range of the inverse function of$\,y=\mathrm{cos}x,f\left(x\right)={\mathrm{cos}}^{-1}x.\,$

Since the functions$\,y=\mathrm{cos}\,x\,$and$\,y={\mathrm{cos}}^{-1}x\,$are inverse functions, why is$\,{\mathrm{cos}}^{-1}\left(\mathrm{cos}\left(-\frac{\pi }{6}\right)\right)\,$not equal to$\,-\frac{\pi }{6}?$

Explain the meaning of$\,\frac{\pi }{6}=\mathrm{arcsin}\left(0.5\right).$

Show Solution

$\frac{\pi }{6}\,$is the radian measure of an angle between$\,-\frac{\pi }{2}\,$and$\,\frac{\pi }{2}$whose sine is 0.5.

Most calculators do not have a key to evaluate$\,{\mathrm{sec}}^{-1}\left(2\right).\,$Explain how this can be done using the cosine function or the inverse cosine function.

Why must the domain of the sine function,$\,\mathrm{sin}\,x,\,$be restricted to$\,\left[-\frac{\pi }{2},\frac{\pi }{2}\right]\,$for the inverse sine function to exist?

Show Solution

In order for any function to have an inverse, the function must be one-to-one and must pass the horizontal line test. The regular sine function is not one-to-one unless its domain is restricted in some way. Mathematicians have agreed to restrict the sine function to the interval$\,\left[-\frac{\pi }{2},\frac{\pi }{2}\right]\,$so that it is one-to-one and possesses an inverse.

Discuss why this statement is incorrect:$\,\mathrm{arccos}\left(\mathrm{cos}\,x\right)=x\,$for all$\,x.$

Determine whether the following statement is true or false and explain your answer: $\mathrm{arccos}\left(-x\right)=\pi -\mathrm{arccos}\,x.$

Show Solution

True . The angle,$\,{\theta }_{1}\,$that equals$\,\mathrm{arccos}\left(-x\right)\,$,$\,x>0\,$, will be a second quadrant angle with reference angle,$\,{\theta }_{2}\,$, where$\,{\theta }_{2}\,$equals$\,\mathrm{arccos}x$,$x>0\,$. Since$\,{\theta }_{2}\,$is the reference angle for$\,{\theta }_{1}$,${\theta }_{2}=\pi -{\theta }_{1}\,$and$\,\mathrm{arccos}\left(-x\right)\,$=$\,\pi -\mathrm{arccos}x$–

Algebraic

For the following exercises, evaluate the expressions.

${\mathrm{sin}}^{-1}\left(\frac{\sqrt{2}}{2}\right)$

${\mathrm{sin}}^{-1}\left(-\frac{1}{2}\right)$

Show Solution

$-\frac{\pi }{6}$

${\mathrm{cos}}^{-1}\left(\frac{1}{2}\right)$

${\mathrm{cos}}^{-1}\left(-\frac{\sqrt{2}}{2}\right)$

Show Solution

$\frac{3\pi }{4}$

${\mathrm{tan}}^{-1}\left(1\right)$

${\mathrm{tan}}^{-1}\left(-\sqrt{3}\right)$

Show Solution

$-\frac{\pi }{3}$

${\mathrm{tan}}^{-1}\left(-1\right)$

${\mathrm{tan}}^{-1}\left(\sqrt{3}\right)$

Show Solution

$\frac{\pi }{3}$

${\mathrm{tan}}^{-1}\left(\frac{-1}{\sqrt{3}}\right)$

For the following exercises, use a calculator to evaluate each expression. Express answers to the nearest hundredth.

${\mathrm{cos}}^{-1}\left(-0.4\right)$

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1.98

$\mathrm{arcsin}\left(0.23\right)$

$\mathrm{arccos}\left(\frac{3}{5}\right)$

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0.93

${\mathrm{cos}}^{-1}\left(0.8\right)$

${\mathrm{tan}}^{-1}\left(6\right)$

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1.41

For the following exercises, find the angle$\,\theta \,$in the given right triangle. Round answers to the nearest hundredth.

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0.56 radians

For the following exercises, find the exact value, if possible, without a calculator. If it is not possible, explain why.

${\mathrm{sin}}^{-1}\left(\mathrm{cos}\left(\pi \right)\right)$

${\mathrm{tan}}^{-1}\left(\mathrm{sin}\left(\pi \right)\right)$

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0

${\mathrm{cos}}^{-1}\left(\mathrm{sin}\left(\frac{\pi }{3}\right)\right)$

${\mathrm{tan}}^{-1}\left(\mathrm{sin}\left(\frac{\pi }{3}\right)\right)$

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0.71

${\mathrm{sin}}^{-1}\left(\mathrm{cos}\left(\frac{-\pi }{2}\right)\right)$

${\mathrm{tan}}^{-1}\left(\mathrm{sin}\left(\frac{4\pi }{3}\right)\right)$

Show Solution

-0.71

${\mathrm{sin}}^{-1}\left(\mathrm{sin}\left(\frac{5\pi }{6}\right)\right)$

${\mathrm{tan}}^{-1}\left(\mathrm{sin}\left(\frac{-5\pi }{2}\right)\right)$

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$-\frac{\pi }{4}$

$\mathrm{cos}\left({\mathrm{sin}}^{-1}\left(\frac{4}{5}\right)\right)$

$\mathrm{sin}\left({\mathrm{cos}}^{-1}\left(\frac{3}{5}\right)\right)$

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0.8

$\mathrm{sin}\left({\mathrm{tan}}^{-1}\left(\frac{4}{3}\right)\right)$

$\mathrm{cos}\left({\mathrm{tan}}^{-1}\left(\frac{12}{5}\right)\right)$

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$\frac{5}{13}$

$\mathrm{cos}\left({\mathrm{sin}}^{-1}\left(\frac{1}{2}\right)\right)$

For the following exercises, find the exact value of the expression in terms of$\,x\,$
with the help of a reference triangle.

$\mathrm{tan}\left({\mathrm{sin}}^{-1}\left(x-1\right)\right)$

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$\frac{x-1}{\sqrt{-{x}^{2}+2x}}$

$\mathrm{sin}\left({\mathrm{cos}}^{-1}\left(1-x\right)\right)$

$\mathrm{cos}\left({\mathrm{sin}}^{-1}\left(\frac{1}{x}\right)\right)$

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$\frac{\sqrt{{x}^{2}-1}}{x}$

$\mathrm{cos}\left({\mathrm{tan}}^{-1}\left(3x-1\right)\right)$

$\mathrm{tan}\left({\mathrm{sin}}^{-1}\left(x+\frac{1}{2}\right)\right)$

Show Solution

$\frac{x+0.5}{\sqrt{-{x}^{2}-x+\frac{3}{4}}}$

Extensions

For the following exercises, evaluate the expression without using a calculator. Give the exact value.

$\frac{{\mathrm{sin}}^{-1}\left(\frac{1}{2}\right)-{\mathrm{cos}}^{-1}\left(\frac{\sqrt{2}}{2}\right)+{\mathrm{sin}}^{-1}\left(\frac{\sqrt{3}}{2}\right)-{\mathrm{cos}}^{-1}\left(1\right)}{{\mathrm{cos}}^{-1}\left(\frac{\sqrt{3}}{2}\right)-{\mathrm{sin}}^{-1}\left(\frac{\sqrt{2}}{2}\right)+{\mathrm{cos}}^{-1}\left(\frac{1}{2}\right)-{\mathrm{sin}}^{-1}\left(0\right)}$

For the following exercises, find the function if$\,\mathrm{sin}\,t=\frac{x}{x+1}.$

$\mathrm{cos}\,t$

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$\frac{\sqrt{2x+1}}{x+1}$

$\mathrm{sec}\,t$

$\mathrm{cot}\,t$

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$\frac{\sqrt{2x+1}}{x}$

$\mathrm{cos}\left({\mathrm{sin}}^{-1}\left(\frac{x}{x+1}\right)\right)$

${\mathrm{tan}}^{-1}\left(\frac{x}{\sqrt{2x+1}}\right)$

Show Solution

$t$

Graphical

Graph$\,y={\mathrm{sin}}^{-1}x\,$and state the domain and range of the function.

Graph$\,y=\mathrm{arccos}\,x\,$and state the domain and range of the function.

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domain$\,\left[-1,1\right];\,$range$\,\left[0,\pi \right]\,$

Graph one cycle of$\,y={\mathrm{tan}}^{-1}x\,$and state the domain and range of the function.

For what value of$\,x\,$does$\,\mathrm{sin}\,x={\mathrm{sin}}^{-1}x?\,$Use a graphing calculator to approximate the answer.

Show Solution

approximately$\,x=0.00\,$

For what value of$\,x\,$does$\,\mathrm{cos}\,x={\mathrm{cos}}^{-1}x?\,$Use a graphing calculator to approximate the answer.

Real-World Applications

Suppose a 13-foot ladder is leaning against a building, reaching to the bottom of a second-floor window 12 feet above the ground. What angle, in radians, does the ladder make with the building?

Show Solution

0.395 radians

Suppose you drive 0.6 miles on a road so that the vertical distance changes from 0 to 150 feet. What is the angle of elevation of the road?

An isosceles triangle has two congruent sides of length 9 inches. The remaining side has a length of 8 inches. Find the angle that a side of 9 inches makes with the 8-inch side.

Show Solution

1.11 radians

Without using a calculator, approximate the value of$\,\mathrm{arctan}\left(10,000\right).\,$Explain why your answer is reasonable.

A truss for the roof of a house is constructed from two identical right triangles. Each has a base of 12 feet and height of 4 feet. Find the measure of the acute angle adjacent to the 4-foot side.

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1.25 radians

The line$\,y=\frac{3}{5}x\,$passes through the origin in the x,y-plane. What is the measure of the angle that the line makes with the positive x-axis?

The line$\,y=\frac{-3}{7}x\,$passes through the origin in the x,y-plane. What is the measure of the angle that the line makes with the negative x-axis?

Show Solution

0.405 radians

What percentage grade should a road have if the angle of elevation of the road is 4 degrees? (The percentage grade is defined as the change in the altitude of the road over a 100-foot horizontal distance. For example a 5% grade means that the road rises 5 feet for every 100 feet of horizontal distance.)

A 20-foot ladder leans up against the side of a building so that the foot of the ladder is 10 feet from the base of the building. If specifications call for the ladder’s angle of elevation to be between 35 and 45 degrees, does the placement of this ladder satisfy safety specifications?

Show Solution

No. The angle the ladder makes with the horizontal is 60 degrees.

Suppose a 15-foot ladder leans against the side of a house so that the angle of elevation of the ladder is 42 degrees. How far is the foot of the ladder from the side of the house?

Chapter Review Exercises

Graphs of the Sine and Cosine Functions

For the following exercises, graph the functions for two periods and determine the amplitude or stretching factor, period, midline equation, and asymptotes.

$f\left(x\right)=-3\mathrm{cos}\,x+3$

Show Solution

amplitude: 3; period:$\,2\pi ;\,$midline:$\,y=3;\,$no asymptotes

$f\left(x\right)=\frac{1}{4}\mathrm{sin}\,x$

$f\left(x\right)=3\mathrm{cos}\left(x+\frac{\pi }{6}\right)$

Show Solution

amplitude: 3; period:$\,2\pi ;\,$midline:$\,y=0;\,$no asymptotes

$f\left(x\right)=-2\mathrm{sin}\left(x-\frac{2\pi }{3}\right)$

$f\left(x\right)=3\mathrm{sin}\left(x-\frac{\pi }{4}\right)-4$

Show Solution

amplitude: 3; period:$\,2\pi ;\,$midline:$\,y=-4;\,$no asymptotes

$f\left(x\right)=2\left(\mathrm{cos}\left(x-\frac{4\pi }{3}\right)+1\right)$

$f\left(x\right)=6\mathrm{sin}\left(3x-\frac{\pi }{6}\right)-1$

Show Solution

amplitude: 6; period:$\,\frac{2\pi }{3};\,$midline:$\,y=-1;\,$no asymptotes

$f\left(x\right)=-100\mathrm{sin}\left(50x-20\right)$

Graphs of the Other Trigonometric Functions

For the following exercises, graph the functions for two periods and determine the amplitude or stretching factor, period,
midline equation, and asymptotes.

$f\left(x\right)=\mathrm{tan}\,x-4$

Show Solution

stretching factor: none; period:$\text{ }\pi ;\text{ }$midline:$\text{ }y=-4;\text{ }$asymptotes:$\text{ }x=\frac{\pi }{2}+\pi k,$where$\text{ }k\text{ }$is an integer

$f\left(x\right)=2\mathrm{tan}\left(x-\frac{\pi }{6}\right)$

$f\left(x\right)=-3\mathrm{tan}\left(4x\right)-2$

Show Solution

stretching factor: 3; period:$\text{ }\frac{\pi }{4};\text{ }$midline:$\text{ }y=-2;\text{ }$asymptotes:$x=\frac{\pi }{8}+\frac{\pi }{4}k,$where$\text{ }k\text{ }$is an integer

$f\left(x\right)=0.2\mathrm{cos}\left(0.1x\right)+0.3$

For the following exercises, graph two full periods. Identify the period, the phase shift, the amplitude, and asymptotes.

$f\left(x\right)=\frac{1}{3}\mathrm{sec}\,x$

Show Solution

amplitude: none; period:$2\pi ;$no phase shift; asymptotes:$\text{ }x=\frac{\pi }{2}k,$where$\text{ }k\text{ }$is an odd integer

$f\left(x\right)=3\mathrm{cot}\,x$

$f\left(x\right)=4\mathrm{csc}\left(5x\right)$

Show Solution

amplitude: none; period:$\text{ }\frac{2\pi }{5};\text{ }$no phase shift; asymptotes:$\text{ }x=\frac{\pi }{5}k,$where$\text{ }k\text{ }$is an integer

$f\left(x\right)=8\mathrm{sec}\left(\frac{1}{4}x\right)$

$f\left(x\right)=\frac{2}{3}\mathrm{csc}\left(\frac{1}{2}x\right)$

Show Solution

amplitude: none; period:$\text{ }4\pi ;\text{ }$no phase shift; asymptotes:$\text{ }x=2\pi k,$where$\text{ }k\text{ }$is an integer

$f\left(x\right)=-\mathrm{csc}\left(2x+\pi \right)$

For the following exercises, use this scenario: The population of a city has risen and fallen over a 20-year interval. Its population may be modeled by the following function:$\,y=12,000+8,000\mathrm{sin}\left(0.628x\right),\,$where the domain is the years since 1980 and the range is the population of the city.

What is the largest and smallest population the city may have?

Show Solution

largest: 20,000; smallest: 4,000

Graph the function on the domain of$\,\left[0,40\right]$.

What are the amplitude, period, and phase shift for the function?

Show Solution

amplitude: 8,000; period: 10; phase shift: 0

Over this domain, when does the population reach 18,000? 13,000?

What is the predicted population in 2007? 2010?

Show Solution

In 2007, the predicted population is 4,413. In 2010, the population will be 11,924.

For the following exercises, suppose a weight is attached to a spring and bobs up and down, exhibiting symmetry.

Suppose the graph of the displacement function is shown in (Figure), where the values on the x-axis represent the time in seconds and the y-axis represents the displacement in inches. Give the equation that models the vertical displacement of the weight on the spring.

Figure 13.

At time = 0, what is the displacement of the weight?

Show Solution

5 in.

At what time does the displacement from the equilibrium point equal zero?

What is the time required for the weight to return to its initial height of 5 inches? In other words, what is the period for the displacement function?

Show Solution

10 seconds

Inverse Trigonometric Functions

For the following exercises, find the exact value without the aid of a calculator.

${\mathrm{sin}}^{-1}\left(1\right)$

${\mathrm{cos}}^{-1}\left(\frac{\sqrt{3}}{2}\right)$

Show Solution

$\frac{\pi }{6}$

${\mathrm{tan}}^{-1}\left(-1\right)$

${\mathrm{cos}}^{-1}\left(\frac{1}{\sqrt{2}}\right)$

Show Solution

$\frac{\pi }{4}$

${\mathrm{sin}}^{-1}\left(\frac{-\sqrt{3}}{2}\right)$

${\mathrm{sin}}^{-1}\left(\mathrm{cos}\left(\frac{\pi }{6}\right)\right)$

Show Solution

$\frac{\pi }{3}$

${\mathrm{cos}}^{-1}\left(\mathrm{tan}\left(\frac{3\pi }{4}\right)\right)$

$\mathrm{sin}\left({\mathrm{sec}}^{-1}\left(\frac{3}{5}\right)\right)$

Show Solution

No solution

$\mathrm{cot}\left({\mathrm{sin}}^{-1}\left(\frac{3}{5}\right)\right)$

$\mathrm{tan}\left({\mathrm{cos}}^{-1}\left(\frac{5}{13}\right)\right)$

Show Solution

$\frac{12}{5}$

$\mathrm{sin}\left({\mathrm{cos}}^{-1}\left(\frac{x}{x+1}\right)\right)$

Graph$\,f\left(x\right)=\mathrm{cos}\,x\,$and$\,f\left(x\right)=\mathrm{sec}\,x\,$on the interval$\,\left[0,2\pi \right)\,$and explain any observations.

Show Solution

The graphs are not symmetrical with respect to the line$\,y=x.\,$They are symmetrical with respect to the$\,y$-axis.

Graph$\,f\left(x\right)=\mathrm{sin}\,x\,$and$\,f\left(x\right)=\mathrm{csc}\,x\,$and explain any observations.

Graph the function$f\,\left(x\right)=\frac{x}{1}-\frac{{x}^{3}}{3!}+\frac{{x}^{5}}{5!}-\frac{{x}^{7}}{7!}\,$on the interval$\,\left[-1,1\right]\,$and compare the graph to the graph of$\,f\left(x\right)=\mathrm{sin}\,x\,$on the same interval. Describe any observations.

Show Solution

The graphs appear to be identical.

Chapter Practice Test

For the following exercises, sketch the graph of each function for two full periods. Determine the amplitude, the period, and the equation for the midline.

$f\left(x\right)=0.5\mathrm{sin}\,x$

Show Solution

amplitude: 0.5; period:$\,2\pi ;\,$midline$\,y=0\,$

$f\left(x\right)=5\mathrm{cos}\,x$

$f\left(x\right)=5\mathrm{sin}\,x$

Show Solution

amplitude: 5; period:$\,2\pi ;\,$midline:$\,y=0$

$f\left(x\right)=\mathrm{sin}\left(3x\right)$

$f\left(x\right)=-\mathrm{cos}\left(x+\frac{\pi }{3}\right)+1$

Show Solution

amplitude: 1; period:$\,2\pi ;\,$midline:$\,y=1$

$f\left(x\right)=5\mathrm{sin}\left(3\left(x-\frac{\pi }{6}\right)\right)+4$

$f\left(x\right)=3\mathrm{cos}\left(\frac{1}{3}x-\frac{5\pi }{6}\right)$

Show Solution

amplitude: 3; period:$\,6\pi ;\,$midline:$\,y=0$

$f\left(x\right)=\mathrm{tan}\left(4x\right)$

$f\left(x\right)=-2\mathrm{tan}\left(x-\frac{7\pi }{6}\right)+2$

Show Solution

amplitude: none; period:$\text{ }\pi ;\text{ }$midline:$\text{ }y=0,$asymptotes:$\text{ }x=\frac{2\pi }{3}+\pi k,$where$\text{ }k\text{ }$is an integer

$f\left(x\right)=\pi \mathrm{cos}\left(3x+\pi \right)$

$f\left(x\right)=5\mathrm{csc}\left(3x\right)$

Show Solution

amplitude: none; period:$\text{ }\frac{2\pi }{3};\text{ }$midline:$\text{ }y=0,$ asymptotes:$\text{ }x=\frac{\pi }{3}k,$ where$\text{ }k\text{ }$is an integer

$f\left(x\right)=\pi \mathrm{sec}\left(\frac{\pi }{2}x\right)$

$f\left(x\right)=2\mathrm{csc}\left(x+\frac{\pi }{4}\right)-3$

Show Solution

amplitude: none; period:$\,2\pi ;\,$midline:$\,y=-3$

For the following exercises, determine the amplitude, period, and midline of the graph, and then find a formula for the function.

Give in terms of a sine function.

Give in terms of a sine function.

Show Solution

amplitude: 2; period: 2; midline:$\,y=0;$$f\left(x\right)=2\mathrm{sin}\left(\pi \left(x-1\right)\right)$

Give in terms of a tangent function.

For the following exercises, find the amplitude, period, phase shift, and midline.

$y=\mathrm{sin}\left(\frac{\pi }{6}x+\pi \right)-3$

Show Solution

amplitude: 1; period: 12; phase shift:$\,-6;\,$midline$\,y=-3$

$y=8\mathrm{sin}\left(\frac{7\pi }{6}x+\frac{7\pi }{2}\right)+6$

The outside temperature over the course of a day can be modeled as a sinusoidal function. Suppose you know the temperature is 68°F at midnight and the high and low temperatures during the day are 80°F and 56°F, respectively. Assuming$\,t\,$is the number of hours since midnight, find a function for the temperature,$\,D,\,$in terms of$\,t.$

Show Solution

$D\left(t\right)=68-12\mathrm{sin}\left(\frac{\pi }{12}x\right)$

Water is pumped into a storage bin and empties according to a periodic rate. The depth of the water is 3 feet at its lowest at 2:00 a.m. and 71 feet at its highest, which occurs every 5 hours. Write a cosine function that models the depth of the water as a function of time, and then graph the function for one period.

For the following exercises, find the period and horizontal shift of each function.

$g\left(x\right)=3\mathrm{tan}\left(6x+42\right)$

Show Solution

period:$\,\frac{\pi }{6};\,$horizontal shift:$\,-7$

$n\left(x\right)=4\mathrm{csc}\left(\frac{5\pi }{3}x-\frac{20\pi }{3}\right)$

Write the equation for the graph in (Figure) in terms of the secant function and give the period and phase shift.

Figure 14.

Show Solution

$f\left(x\right)=\mathrm{sec}\left(\pi x\right);\,$period: 2; phase shift: 0

If$\,\mathrm{tan}\,x=3,\,$find$\,\mathrm{tan}\left(-x\right).$

If$\,\mathrm{sec}\,x=4,\,$find$\,\mathrm{sec}\left(-x\right).$

Show Solution

$4$

For the following exercises, graph the functions on the specified window and answer the questions.

Graph$\,m\left(x\right)=\mathrm{sin}\left(2x\right)+\mathrm{cos}\left(3x\right)\,$on the viewing window$\,\left[-10,10\right]\,$by$\,\left[-3,3\right].\,$Approximate the graph’s period.

Graph$\,n\left(x\right)=0.02\mathrm{sin}\left(50\pi x\right)\,$on the following domains in$\,x:$$\left[0,1\right]\,$and$\,\left[0,3\right].\,$Suppose this function models sound waves. Why would these views look so different?

Show Solution

The views are different because the period of the wave is$\,\frac{1}{25}.\,$Over a bigger domain, there will be more cycles of the graph.

Graph$\,f\left(x\right)=\frac{\mathrm{sin}\,x}{x}\,$on$\,\left[-0.5,0.5\right]\,$and explain any observations.

For the following exercises, let$\,f\left(x\right)=\frac{3}{5}\mathrm{cos}\left(6x\right).$

What is the largest possible value for$\,f\left(x\right)?$

Show Solution

$\frac{3}{5}$

What is the smallest possible value for$\,f\left(x\right)?$

Where is the function increasing on the interval$\,\left[0,2\pi \right]?$

Show Solution

On the approximate intervals$\,\left(0.5,1\right),\left(1.6,2.1\right),\left(2.6,3.1\right),\left(3.7,4.2\right),\left(4.7,5.2\right),\left(5.6,6.28\right)$

For the following exercises, find and graph one period of the periodic function with the given amplitude, period, and phase shift.

Sine curve with amplitude 3, period$\,\frac{\pi }{3},\,$and phase shift$\,\left(h,k\right)=\left(\frac{\pi }{4},2\right)$

Cosine curve with amplitude 2, period$\,\frac{\pi }{6},\,$and phase shift$\,\left(h,k\right)=\left(-\frac{\pi }{4},3\right)$

Show Solution

$f\left(x\right)=2\mathrm{cos}\left(12\left(x+\frac{\pi }{4}\right)\right)+3$

For the following exercises, graph the function. Describe the graph and, wherever applicable, any periodic behavior, amplitude, asymptotes, or undefined points.

$f\left(x\right)=5\mathrm{cos}\left(3x\right)+4\mathrm{sin}\left(2x\right)$

$f\left(x\right)={e}^{\mathrm{sin}t}$

Show Solution

This graph is periodic with a period of$\,2\pi .$

For the following exercises, find the exact value.

${\mathrm{sin}}^{-1}\left(\frac{\sqrt{3}}{2}\right)$

${\mathrm{tan}}^{-1}\left(\sqrt{3}\right)$

Show Solution

$\frac{\pi }{3}$

${\mathrm{cos}}^{-1}\left(-\frac{\sqrt{3}}{2}\right)$

${\mathrm{cos}}^{-1}\left(\mathrm{sin}\left(\pi \right)\right)$

Show Solution

$\frac{\pi }{2}$

${\mathrm{cos}}^{-1}\left(\mathrm{tan}\left(\frac{7\pi }{4}\right)\right)$

$\mathrm{cos}\left({\mathrm{sin}}^{-1}\left(1-2x\right)\right)$

Show Solution

$\sqrt{1-{\left(1-2x\right)}^{2}}$

${\mathrm{cos}}^{-1}\left(-0.4\right)$

$\mathrm{cos}\left({\mathrm{tan}}^{-1}\left({x}^{2}\right)\right)$

Show Solution

$\frac{1}{\sqrt{1+{x}^{4}}}$

For the following exercises, suppose$\,\mathrm{sin}\,t=\frac{x}{x+1}.$
Evaluate the following expressions.

$\mathrm{tan}\,t$

$\mathrm{csc}\,t$

Show Solution

$\frac{x+1}{x}$

Given (Figure), find the measure of angle$\,\theta \,$to three decimal places. Answer in radians.

Figure 15.

For the following exercises, determine whether the equation is true or false.

$\mathrm{arcsin}\left(\mathrm{sin}\left(\frac{5\pi }{6}\right)\right)=\frac{5\pi }{6}$

Show Solution

False

$\mathrm{arccos}\left(\mathrm{cos}\left(\frac{5\pi }{6}\right)\right)=\frac{5\pi }{6}$

The grade of a road is 7%. This means that for every horizontal distance of 100 feet on the road, the vertical rise is 7 feet. Find the angle the road makes with the horizontal in radians.

Show Solution

approximately 0.07 radians

Glossary

arccosine

another name for the inverse cosine;$\,\mathrm{arccos}\,x={\mathrm{cos}}^{-1}x$

arcsine

another name for the inverse sine;$\,\mathrm{arcsin}\,x={\mathrm{sin}}^{-1}x$

arctangent

another name for the inverse tangent;$\,\mathrm{arctan}\,x={\mathrm{tan}}^{-1}x$

inverse cosine function

the function$\,{\mathrm{cos}}^{-1}x,\,$which is the inverse of the cosine function and the angle that has a cosine equal to a given number

inverse sine function

the function$\,{\mathrm{sin}}^{-1}x,\,$which is the inverse of the sine function and the angle that has a sine equal to a given number

inverse tangent function

the function $\,{\mathrm{tan}}^{-1}x,\,$which is the inverse of the tangent function and the angle that has a tangent equal to a given number