Section 7.1 Inverse Trigonometric Functions

Transcription

1 6 Section 7.1 Inverse Trigonometric Functions Recall that if we examine function like p(x) = x + 4, it was not one-to-one if there was a horizontal line that passed through more than one point on the graph: Since it is not one-to-one, it does not have an inverse function. But, if we restricted the domain of f so that you could not find a horizontal line that passes through more than one point on the graph, then the function with that restricted domain would be one-to-one and have an inverse function. Consider p(x) = x + 4, x 0 This function has a restricted domain, giving only the left half of the parabola. By the horizontal test, the function with this restricted domain is one-to-one. The domain is {x x 0} and the range is {y y 4}. The inverse function is p 1 (x) = x 4. We have also seen that the graph of a function and its inverse are symmetrical to the line y = x.

2 64 p p 1 If we examine the function f(x) = sin(x), we can see that it is not one-to-one. 5 However, suppose we restrict the domain to [, ]: 5 Thus, the sine function with this domain will have an inverse function. In our definition of the inverse sine function, we reverse the roles of x and y.

3 65 Defintion Let x = sin(y) be a function on the restricted domain of [, ]. Then, the inverse sine function (arcsine) is defined as: y = sin 1 (x) or arcsin(x) where 1 x 1 and y. The value y is the angle between and inclusively whose sine is x. To obtain the graph of sin 1 (x), we can reflect the graph of sin(x) defined on the interval [, ] across the y = x line. sin 1 (x) sin(x) Objective 1: Find the Exact Value of an Inverse Sine Function. Find the Exact Value of the following: Ex. 1a sin 1 ( 1) Ex. 1b sin 1 ( ) a) We are looking for a value of y in [, ] such that sin(y) = 1. The sine is equal to 1 when the terminal side lies on the negative y-axis. Thus, the angle is. b) We are looking for a value of y in [, ] such that sin(y) =. The sine is equal to in quadrant I. Thus, the angle is. when the terminal side is

4 66 Objective : Find an Approximate Value of an Inverse Sine Function. On most calculators, you will need to hit the "Shift" or "nd" key followed by the "SIN" key to obtain the answer. Find the following. Round to two decimal places: Ex. a sin 1 ( 5 ) Ex. b sin 1 ( 0.75) Be sure you calculator is in radian mode, before you begin. a) sin 1 ( ) = radians. 5 b) sin 1 ( 0.75) = radians. Objective : Using Properties of Inverse Functions to Find the Exact Value of Certain Composite Functions. Recall that in our discussion of inverse functions, if we were finding (f o g)(x) where f and g are inverse functions, the value of x had to be in the domain of the inside function. Applying this to the sine and inverse sine functions, we get: 1) sin 1 (sin(x)) = x where x. ) sin(sin 1 (x)) = x where 1 x 1. Note, if x is not between and inclusively, the composition sin 1 (sin(x)) = x will produce an angle that is between and inclusively. So, we get the reference angle if sin(x) > 0, or the opposite of the reference angle if sin(x) < 0. Find the Exact Value of the Following: Ex. a sin 1 (sin( 7 )) Ex. b sin 1 (sin( 9 5 )) Ex. c sin(sin 1 (0.6)) Ex. d sin(sin 1 ( )) a) Since 7 is in the interval in [, ], then sin 1 (sin( 7 )) = 7. b) 9 5 is in quadrant IV, so the reference angle is 9 5 = 5. Since sin( 9 5 ) is negative, then sin 1 (sin( 9 5 )) = 5

5 67 c) Since 0.6 is in the interval [ 1, 1], then sin(sin 1 (0.6)) = 0.6. d) Since is not in the domain of the inverse sine, then sin(sin 1 ( )) is undefined. If we examine the graph of the cosine function, we see that we will have the same problem that we had with the sine function: 5 Clearly, it is not one-to-one. So, we will need to restrict the domain. This time, we will choose the interval [0, ]: 5 Thus, the cosine function with this domain will have an inverse function. In our definition of the inverse cosine function, we reverse the roles of x and y.

6 Defintion Let x = cosine(y) be a function on the restricted domain of [0, ]. Then, the inverse cosine function (arccosine) is defined as: y = cos 1 (x) or arccos(x) where 1 x 1 and 0 y. The value y is the angle between 0 and inclusively whose cosine is x. To obtain the graph of cos 1 (x), we can reflect the graph of cos(x) defined on the interval [0, ] across the y = x line. 68 cos 1 (x) cos(x) Find the Exact Value of the following: Ex. 4a cos 1 ( 1) Ex. 4b cos 1 ( ) a) We are looking for a value of y in [0, ] such that cos(y) = 1. The cosine is equal to 1 when the terminal side lies on the negative x-axis. Thus, the angle is. b) We are looking for a value of y in [0, ]. such that cos(y) =. The cosine is equal to when the terminal side is in quadrant I. Thus, the angle is 6. On most calculators, you will need to hit the "Shift" or "nd" key followed by the "COS" key to obtain the answer. Find the following. Round to two decimal places: Ex. 5a cos 1 ( 5 ) Ex. 5b cos 1 ( 0.75)

7 69 Be sure you calculator is in radian mode, before you begin. a) cos 1 ( ) = radians. 5 b) cos 1 ( 0.75) = radians. Applying the inverse properties with composition of the cosine and inverse cosine functions, we get: 1) cos 1 (cos(x)) = x where 0 x. ) cos(cos 1 (x)) = x where 1 x 1. Note, if x is not between 0 and inclusively, the composition cos 1 (cos(x)) = x will produce an angle that is between 0 and inclusively. So, we get the reference angle if cos(x) > 0, or minus the reference angle if cos(x) < 0. Find the Exact Value of the Following: Ex. 6a cos 1 (cos( 8 )) Ex. 6b cos 1 (cos( 8 7 )) Ex. 6c cos(cos 1 ( 0.78)) Ex. 6d cos(cos 1 ( )) a) Since 8 is in the interval in [0, ] then cos 1 (cos( 8 )) = 8. b) 8 7 is in quadrant III, so the reference angle is 8 7 = 7. Since cos( 8 7 ) is negative, then cos 1 (cos( 8 )) will yield an 7 angle in quadrant II which is 7 = 6 7. Hence, cos 1 (cos( 8 7 )) = 6 7 c) Since 0.78 is in the interval [ 1, 1], then cos(cos 1 ( 0.78)) = d) Since cos(cos 1 ( is not in the domain of the inverse cosine, then )) is undefined.

8 70 If we examine the function f(x) = tan(x), we can see that it is not one-toone. However, suppose we restrict the domain to (, ): Thus, the tangent function with this domain will have an inverse function. In our definition of the inverse tangent function, we reverse the roles of x & y. Defintion Let x = tan(y) be a function on the restricted domain of (, ). Then, the inverse tangent function (arctangent) is defined as: y = tan 1 (x) or arctan(x) where < x < and < y <.

9 71 The value y is the angle between and inclusively whose tangent is x. To obtain the graph of tan 1 (x), we can reflect the graph of tan(x) defined on the interval (, ) across the y = x line. tan(x) tan 1 (x) Find the Exact Value of the following: Ex. 7a tan 1 (1) Ex. 7b tan 1 (0) a) We are looking for a value of y in (, ) such that tan(y) = 1. The tangent is equal to 1 when the terminal side lies in the first quadrant. Thus, the angle is 4. b) We are looking for a value of y in (, ) such that tan(y) = 0. The tangent is equal to 0 when the terminal side lies on the x-axis. Thus, the angle is 0. On most calculators, you will need to hit the "Shift" or "nd" key followed by the "TAN" key to obtain the answer. Find the following. Round to two decimal places: Ex. 8a tan 1 ( 5 8 ) Ex. 8b tan 1 (.) Be sure you calculator is in radian mode, before you begin.

10 7 a) tan 1 ( 5 ) = radians. 8 b) tan 1 (.) = radians. Applying the inverse properties with composition of the tangent and inverse tangent function, we get: 1) tan 1 (tan(x)) = x where < x <. ) tan(tan 1 (x)) = x where < x <. Note, if x is not between and inclusively, the composition tan 1 (tan(x)) = x will produce an angle that is between and inclusively. So, we get the reference angle if tan(x) > 0, or the opposite of the reference angle if tan(x) < 0. Find the Exact Value of the Following: Ex. 9a tan 1 (tan( 5 )) Ex. 9b tan 1 (tan( 11 5 )) Ex. 9c tan(tan 1 (4.67)) Ex. 9d tan(tan 1 ( )) a) Since 5 is in the interval in (, ), then tan 1 (tan( ))= 5 5. b) 11 5 is in quadrant III, so the reference angle is 11 5 = 5. Since tan( 11 5 ) is positive, then tan 1 (tan( 11 5 )) = 5 c) Since 4.67 is in the interval (, ), then tan(tan 1 (4.67)) = 4.67 d) Since is in the interval (, ), then tan(tan 1 ( )) =. Objective 4: Finding the Inverse Function of a Trigonometric Function. To find the inverse function algebraically, we interchange x and y and solve for y. We can do the same thing for trigonometric functions. Find the inverse function of the following. State the domain of the function and the inverse function: Ex. 10a f(x) = cos(x + 1) + 4 Ex. 10b g(x) = tan(x + )

11 a) Domain of f is (, ). Replace f(x) by y: y = cos(x + 1) + 4 (interchange x and y) x = cos(y + 1) + 4 (solve for the cosine function) x 4 = cos(y + 1) x 4 = cos(y + 1) (use the defintion of the inverse cosine) y + 1 = cos 1 x 4 ( ) (solve for y) y = cos 1 x 4 ( ) 1 y = cos 1 x 4 1 cos 1 x 4 1 So, f 1 (x) = To find the domain of f 1, the argument for inverse cosine has to be between 1 and 1 inclusively: 1 x 4 1 (now, solve for x) x 4 x 6 So, the domain of f 1 is [, 6]. b) The argument of the tangent function cannot be equal to odd multiples of (k+1). Thus, x + where k is an integer or x (k+1) where k is an integer. Thus, the domain of f is {x x (k+1), k is an integer} Replace g(x) by y: y = tan(x + ) (interchange x and y) x = tan(y + ) (solve for the tangent function) x + = tan(y + ) x = tan(y + ) (use the defintion of the inverse tangent) y + = tan 1 ( x ) (solve for y) y = tan 1 ( x ) The domain of f 1 is all real numbers. 7

12 74 Objective 5: Solve Equations Involving Inverse Trigonometric Functions. If an equation has an inverse trigonometric function in it, we will need to first isolate the inverse trigonometric function and then apply the definition of the appropriate inverse trigonometric function. Solve the following: Ex. 11a 6cos 1 (x) = 11 Ex. 11b 5 + tan 1 (x) = 4 tan 1 (x) a) First, solve for the inverse cosine: 6cos 1 (x) = 11 cos 1 (x) = 11 (now, apply the defintion) 6 x = cos( 11 6 ) The angle is in quadrant IV, hence the cosine is positive and the reference angle is 11 6 x = cos( 11 6 ) x = x = 6 = 6 (solve for x) The solution is { b) First, solve for the inverse tangent: 5 + tan 1 (x) = 4 tan 1 (x) 5 + 4tan 1 (x) = 4 4tan 1 (x) = tan 1 (x) = 4 x = tan( 4 ). So, cos( 11 6 ) = cos( 6 ) =. 6 }. (now, apply the definition) The angle is in quadrant IV and so, the tangent function is negative. The reference angle is 4. Thus, tan( 4 ) = tan( 4 ) = 1. x = tan( 4 ) x = 1 The solution is { 1}.