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# 6.6

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## 6.6 Inverse Trigonometric

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### 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,

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### 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–1x 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.

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### 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,

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### 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–1x = 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–1x. Then sin y = x and –π/2 ≤ y ≤ π/2.

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

and

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### 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(x2 – 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 ≤ x2 – 1 ≤ 1} = {x | 0 ≤ x2 ≤ 2}

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### Example 2 – Solution

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

(c) The domain of f′ is

{x | –1 < x2 – 1 < 1} = {x | 0 < x2 < 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

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### 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–1x = 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.

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### 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–1x).

Solution1:

Let y = tan–1x. 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:

sec2y = 1 + tan2y

= 1 + x2

Thus

(since sec y > 0 for –π/2 < y < π/2)

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### Example 3 – Solution 2

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

If y = tan–1x, 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–1x = arctan, has domain and range (–π/2, π/2). Its graph is shown in Figure 10.

Figure 10

y = tan–1x = arctan x

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### 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:

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### Example 4

Evaluate arctan Solution:

If we let t = 1/(x – 2), we know that tas 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–1x.

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

and so

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### 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.

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### 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)

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### Example 7

Find

Solution:

If we write

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

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### 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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### Example 9

Find

Solution:

We substitute u = x2 because then du = 2x dx and we can use Equation 14 with a = 3:

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