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**6.6** Inverse Trigonometric

### Functions

### Inverse Trigonometric Functions

*You can see from Figure 1 that the sine function y = sin x is *
not one-to-one (use the Horizontal Line Test).

**Figure 1**

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### Inverse Trigonometric Functions

*But the function f(x) = sin x, –*π/2 ≤ x ≤ π*/2 is one-to-one *
(see Figure 2).The inverse function of this restricted sine

*function f exists and is denoted by sin*^{–1 }or arcsin. It is called
**the inverse sine function or the arcsine function.**

**Figure 2**

*y = sin x,*

### Inverse Trigonometric Functions

Since the definition of an inverse function says that
*f*^{–1}*(x) = y f(y) = x*

we have

Thus, if –1 ≤ x ≤ 1, sin^{–1}*x is the number between –*π/2 and
π*/2 whose sine is x.*

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### Example 1

Evaluate (a) sin^{–1} and (b) tan(arcsin ).

Solution:

(a) We have

because sin(π/6) = and π/6 lies between –π/2 and π/2.

*Example 1 – Solution*

(b) Let θ = arcsin , so sin θ = . Then we can draw a right triangle with angle θ as in Figure 3 and deduce from the Pythagorean Theorem that the third side has length

This enables us to read from the triangle that

**Figure 3**

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### Inverse Trigonometric Functions

The cancellation equations for inverse functions become, in this case,

### Inverse Trigonometric Functions

The inverse sine function, sin^{–1}, has domain [–1, 1] and
range [–π/2, π/2], and its graph, shown in Figure 4, is

obtained from that of the restricted sine function (Figure 2)
*by reflection about the line y = x.*

**Figure 4**

*y = sin*^{–1}*x = arcsin x*

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### Inverse Trigonometric Functions

*We know that the sine function f is continuous, so the *

inverse sine function is also continuous. The sine function is differentiable, so the inverse sine function is also

differentiable.

*Let y = sin*^{–1}*x. Then sin y = x and –*π/2 ≤ y ≤ π/2.

*Differentiating sin y = x implicitly with respect to x, we *
obtain

and

### Inverse Trigonometric Functions

*Now cos y* ≥ 0 since –π/2 ≤ y ≤ π/2, so

Therefore

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### Example 2

*If f(x) = sin*^{–1}*(x*^{2} *– 1), find (a) the domain of f, (b) f′(x), and *
*(c) the domain of f*′.

Solution:

(a) Since the domain of the inverse sine function is [–1, 1],
*the domain of f is*

*{x | –1 ≤ x*^{2} – 1 ≤ 1} = {x | 0 ≤ x^{2} ≤ 2}

*Example 2 – Solution*

(b) Combining Formula 3 with the Chain Rule, we have

*(c) The domain of f*′ is

*{x | –1 < x*^{2} *– 1 < 1} = {x | 0 < x*^{2} < 2}

cont’d

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### Inverse Trigonometric Functions

**The inverse cosine function is handled similarly. The **
*restricted cosine function f(x) = cos x, 0 ≤ x ≤* π, is

one-to-one (see Figure 6) and so it has an inverse function
denoted by cos^{–1} or arccos.

**Figure 6**

*y = cos x, 0 **≤ x ≤ π*

### Inverse Trigonometric Functions

The cancellation equations are

The inverse cosine function, cos^{–1},
has domain [–1, 1] and range [0, π]
and is a continuous function whose
graph is shown in Figure 7.

**Figure 7**

*y = cos*^{–1}*x = arccos x*

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### Inverse Trigonometric Functions

Its derivative is given by

The tangent function can be made one-to-one by restricting it to the interval (–π/2 , π/2).

**Thus the inverse tangent function is defined as the **
*inverse of the function f(x) = tan x, –*π*/2 < x <* π/2.

### Inverse Trigonometric Functions

It is denoted by tan^{–1} or arctan. (See Figure 8.)

**Figure 8**

*y = tan x, < x <*

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### Example 3

Simplify the expression cos(tan^{–1}*x).*

Solution1:

*Let y = tan*^{–1}*x. Then tan y = x and –*π*/2 < y <* π/2. We want
*to find cos y but, since tan y is known, it is easier to find *
*sec y first:*

sec^{2}*y = 1 + tan*^{2}*y*

*= 1 + x*^{2}

Thus

*(since sec y > 0 for –*π*/2 < y <* π/2)

*Example 3 – Solution 2*

Instead of using trigonometric identities as in Solution 1, it is perhaps easier to use a diagram.

*If y = tan*^{–1}*x, then tan y = x, and we can read from Figure 9 *
*(which illustrates the case y > 0) that*

cont’d

**Figure 9**

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### Inverse Trigonometric Functions

The inverse tangent function, tan^{–1}*x = arctan, has domain *
and range (–π/2, π/2). Its graph is shown in Figure 10.

**Figure 10**

*y = tan*^{–1}*x = arctan x*

### Inverse Trigonometric Functions

We know that

and

*and so the lines x = ±*π/2 are vertical asymptotes of the
graph of tan.

Since the graph of tan^{–1 }is obtained by reflecting the graph
*of the restricted tangent function about the line y = x, it *

*follows that the lines y = *π*/2 and y = –*π/2 are horizontal
asymptotes of the graph of tan^{–1}.

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### Inverse Trigonometric Functions

This fact is expressed by the following limits:

### Example 4

Evaluate arctan Solution:

*If we let t = 1/(x – 2), we know that t* → *as x → 2*^{+}.
Therefore, by the first equation in (8), we have

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### Inverse Trigonometric Functions

Because tan is differentiable, tan^{–1} is also differentiable. To
*find its derivative, we let y = tan*^{–1}*x. *

*Then tan y = x. Differentiating this latter equation implicitly *
*with respect to x, we have*

and so

### Inverse Trigonometric Functions

The remaining inverse trigonometric functions are not used as frequently and are summarized here.

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### Inverse Trigonometric Functions

We collect in Table 11 the differentiation formulas for all of the inverse trigonometric functions.

### Inverse Trigonometric Functions

Each of these formulas can be combined with the Chain
*Rule. For instance, if u is a differentiable function of x, then*

and

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### Example 5

*Differentiate (a) y = and (b) f(x) = x arctan .*
Solution:

(a)

(b)

### Inverse Trigonometric Functions

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### Example 7

Find

Solution:

If we write

then the integral resembles Equation 12 and the
*substitution u = 2x is suggested.*

*Example 7 – Solution*

*This gives du = 2 dx, so dx = du /2. When x = 0, u = 0; *

*when x = , u = . So*

cont’d

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### Inverse Trigonometric Functions

### Example 9

Find

Solution:

*We substitute u = x*^{2} *because then du = 2x dx and we can *
*use Equation 14 with a = 3:*