MTH 112: Elementary Functions

Transcription

1 1/19 MTH 11: Elementary Functions Section 6.6

2 6.6:Inverse Trigonometric functions /19 Inverse Trig functions 1 1 functions satisfy the horizontal line test: Any horizontal line crosses the graph of a 1 1 function at most once. f (x) = sin(x) fails to satisfy the horizontal line test. sin(x) 1-1

3 6.6:Inverse Trigonometric functions 3/19 Inverse Trig functions If we restrict the sin function to the interval horizontal line test will indeed be satisfied. [, ], then the sin(x) 1-1

4 6.6:Inverse Trigonometric functions 4/19 Definition of the inverse sine function Definition 1 The inverse sine function, denote by arcsin or sin 1, is defined by y = arcsin(x) if and only if sin(y) = x where 1 x 1 and y. The domain of g(x) = arcsin(x) is [ [ 1,1] and the range is, ]. arcsin(x) 1 sin(x) 1 1 1

5 6.6:Inverse Trigonometric functions 5/19 Example Find the exact value of the given quantity: ( ) 1 1. sin 1 ( ) 3. arcsin 3. sin 1 ( 3)

6 6.6:Inverse Trigonometric functions 6/19 Definition of inverse trigonometric functions Both cosine and tangent functions fail to be 1 1 and, thus, have no inverse. Just as we did with the sine function, this situation can be remedied by restricting the domain to appropriate intervals.

7 6.6:Inverse Trigonometric functions 7/19 arccos(x) cos(x) Definition y = arccos(x) (or y = cos 1 (x)) if and only if cos(y) = x and 0 y.

8 6.6:Inverse Trigonometric functions 8/19 tan(x) arctan(x) Definition 3 y = arctan(x) (or y = tan 1 (x)) if and only if tan(y) = x and < y <.

9 6.6:Inverse Trigonometric functions 9/19 Reviewing: Domain and Range Function Domain Range y = arcsin(x) [ 1,1] [, ] y = arccos(x) [ 1,1] [0,] y = arctan(x) (, ) (, )

10 6.6:Inverse Trigonometric functions 10/19 Example Find the exact value of the given quantity: cos 1 (1) ( ) arccos tan 1 ( 3) arctan( 1)

11 6.6:Inverse Trigonometric functions 11/19 Composing a function and its inverse A function and its inverse undo one another If f is a 1 1 function, so f 1 exits, then for appropriate values of x, f 1 (f (x)) = x and f (f 1 (x)) = x If f (x) = sin(x) is restricted to [, ], then f 1 (x) = arcsin(x), and we have and arcsin(sin(x)) = x, provided x sin(arcsin(x)) = x, provided 1 x 1

12 6.6:Inverse Trigonometric functions 1/19 Summary Property Restriction sin(sin 1 (x)) = x 1 x 1 sin 1 (sin(x)) = x x cos(cos 1 (x)) = x 1 x 1 cos 1 (cos(x)) = x tan(tan 1 (x)) = x 0 x None tan 1 (tan(x)) = x < x <

13 6.6:Inverse Trigonometric functions 13/19 Examples: Composing with the inverse Example 1: Composing with the inverse Evaluate the following quantities: ( ( )) 3 1. cos cos 1 4 ( ( )). arctan tan ( 7 )) 1 3. sin (sin 1 13

14 6.6:Inverse Trigonometric functions 14/19 Solving trigonometric equations with the inverse sine function Example Estimate all solutions of 3sin(x) = with x [0,]. We cannot find an angle x with sine equal to /3 in our trig circle with common angles. We use the sin 1 key on a calculator (set in radian mode) and get x = sin 1 (/3) There is another angle between and with a sine of /3. sin 1 (/3) =

15 6.6:Inverse Trigonometric functions 15/19 Simplifying an expression involving the inverse sine function Example 3 Simplify sin(arccos(x)). Let θ = arccos(x). From the definition of inverse cosine function cos(θ) = x and 0 θ. So, another way of phrasing our problem is: Find a simplified expression for sin(θ) given that cos(θ) = x and 0 θ.

16 6.6:Inverse Trigonometric functions 16/19 Example continues... One way to solve this: When θ is an acute angle: Sketch a triangle having an angle θ whose cosine is x. By the Pythagorian theorem: cos(θ) = x = x 1 opp side = = adj. side hyp. 1 x. Note: We choose positive root because we are dealing with an acute angle. Using the right triangle we have sketched we obtain: sin(θ) = opp side hyp = 1 x 1 = 1 x. We can conclude that sin(arccos(x)) = 1 x.

17 6.6:Inverse Trigonometric functions 17/19 Another way Recall the Pythagorean identity sin (θ) + cos (θ) = 1 and solve for sin(θ). This gives: sin(θ) = ± 1 x. (x = cos(θ)) Since 0 θ then sin(θ) 0. So sin(arccos(x)) = sin(θ) = 1 x.

18 6.6:Inverse Trigonometric functions 18/19 Solving a triangle Example 4 Solve a right triangle with angle β = 63 and side b = 8. Sketch this triangle. Note that α + β + 90 = 180. So α = 7. On the other hand, a b = 8 Then, c = sin(63 ) = 8 c 8 sin(63 ) β = 63 α c

19 6.6:Inverse Trigonometric functions 19/19 Solving a triangle Example 4 Solve a right triangle with angle β = 63 and side b = 8. tan(63 ) = 8 a a b = 8 Then, a = 8 tan(63 ) Can you remember other ways to get a? β = 63 α = 7 c