Inverse Trigonometric Functions Help (page 2)
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
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.
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.
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
Calculate the following derivatives:
You Try It : Calculate ( d/dx )Cos −1 [ x 2 + x ]. Also calculate ( d/dx ) Sin −1 × [ln x − x 3 ].
Calculate each of the following derivatives:
You Try It : Calculate ( d/dx ) ln[Cos −1 x ] and ( d/dx ) exp[Sin −1 x ].
The Inverse Tangent Function
Define the function Tan x to be the restriction of tan x to the interval (− /2, /2). Observe that the tangent function is undefined at the endpoints of this interval. Since
we see that Tan x is increasing, hence it is one-to-one (Fig. 6.19). Also Tan takes arbitrarily large positive values when x is near to, but less than, /2. And Tan takes negative values that are arbitrarily large in absolute value when x is near to, but greater than, −π /2. Therefore Tan takes all real values. Since Tan : (− /2, /2) → (−∞, ∞) is one-to-one and onto, the inverse function Tan −1 : (−∞, ∞) → (− /2, /2) exists. The graph of this inverse function is shown in Fig. 6.20. It is obtained by the usual procedure of reflecting in the line y = x .
As with the first two trigonometric functions, we note that the tangent function takes each of the values at many different points of its domain. But Tan x takes each of these values at just one point of its domain.
The derivative of our new function may be calculated in the usual way. The result is
Next we calculate some derivatives:
Calculate the following derivatives:
You Try It : Calculate ( d/dx ) Tan −1 [ln x + x 3 ] and ( d/dx ) ln[Tan −1 x ].
Integrals in Which Inverse Trigonometric Functions Arise
Our differentiation formulas for inverse trigonometric functions can be written in reverse, as antidifferentiation formulas. We have
The important lesson here is that, while the integrands involve only polynomials and roots, the antiderivatives involve inverse trigonometric functions.
Evaluate the integral
For clarity we set φ( x ) = cos x, φ′ ( x ) = − sin x . The integral becomes
By what we have just learned about Tan −1 , this last integral is equal to
−Tan −1 φ( x ) + C .
Resubstituting φ( x ) = cos x yields that
You Try It : Calculate ∫ x /(1 + x 4 ) dx .
Calculate the integral
For clarity we set φ( x ) = x 3 , φ′( x ) = 3 x 2 . The integral then becomes
We know that this last integral equals
Sin −1 φ ( x ) + C .
Resubstituting the formula for φ gives a final answer of
You Try It : Evaluate the integral
Find practice problems and solutions for these concepts at: Transcendental Functions Practice Test.
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