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# Inverse Trigonometric Functions Help

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By — McGraw-Hill Professional
Updated on Sep 12, 2011

## Introduction to Inverse Trigonometric Functions

Figure 6.14 shows the graphs of each of the six trigonometric functions. Notice that each graph has the property that some horizontal line intersects the graph at least twice. Therefore none of these functions is invertible. Another way of seeing this point is that each of the trigonometric functions is 2π-periodic (that is, the function repeats itself every 2π units: f ( x + 2π) = f ( x )), hence each of these functions is not one-to-one.

If we want to discuss inverses for the trigonometric functions, then we must restrict their domains (this concept was introduced in Subsection 1.8.5). In this section we learn the standard methods for performing this restriction operation with the trigonometric functions

### Inverse Sine Function

Consider the sine function with domain restricted to the interval [− π/2, π/2] (Fig. 6.15). We use the notation Sin x to denote this restricted function. Observe that

Fig. 6.14

Fig. 6.15

on the interval (− π/2, π/2). At the endpoints of the interval, and only there, the function Sin x takes the values −1 and +1. Therefore Sin x is increasing on its entire domain. So it is one-to-one. Furthermore the Sine function assumes every value in the interval [−1, 1]. Thus Sin : [− π/2, π/2] → [−1, 1] is one-to-one and onto; therefore f ( x ) = Sin x is an invertible function.

We can obtain the graph of Sin −1 x by the principle of reflection in the line y = x (Fig. 6.16). The function Sin −1 : [−1, 1] → [−π/2, π/2] is increasing, one-to-one, and onto.

Fig. 6.16

### Inverse Cosine Function

The study of the inverse of cosine involves similar considerations, but we must select a different domain for our function. We define Cos x to be the cosine function restricted to the interval [0, ]. Then, as Fig. 6.17 shows, g ( x ) = Cos x is a one-to-one function. It takes on all the values in the interval [−1, 1]. Thus Cos : [0, ] → [−1, 1] is one-to-one and onto; therefore it possesses an inverse.

We reflect the graph of Cos x in the line y = x to obtain the graph of the function Cos −1 . The result is shown in Fig. 6.18.

Calculate

Fig. 6.17

Fig. 6.18

#### Solution 1

We have

Notice that even though the sine function takes the value at many different values of the variable x , the function Sine takes this value only at x = π/3. Similar comments apply to the other two examples.

We also have

We calculate the derivative of f ( t ) = Sin −1 t by using the usual trick for inverse functions. The result is

The derivative of the function Cos −1 t is calculated much like that of Sin −1 t. We find that

#### Example 2

Calculate the following derivatives:

#### Solution 2

We have

You Try It : Calculate ( d/dx )Cos −1 [ x 2 + x ]. Also calculate ( d/dx ) Sin −1 × [ln x − x 3 ].

#### Example 3

Calculate each of the following derivatives:

#### Solution 3

We have

You Try It : Calculate ( d/dx ) ln[Cos −1 x ] and ( d/dx ) exp[Sin −1 x ].

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