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- 1. The Inverse Trigonometric Functions
- 2. A function is one-to-one (1–1) if different inputs produce different outputs. The Inverse Trigonometric Functions
- 3. A function is one-to-one (1–1) if different inputs produce different outputs. That is, f(x) is 1–1 iff for every pair of inputs u v it must be that f(u) f(v). The Inverse Trigonometric Functions
- 4. A function is one-to-one (1–1) if different inputs produce different outputs. That is, f(x) is 1–1 iff for every pair of inputs u v it must be that f(u) f(v). u f(u) v f(v) u = v f(u) = f(v) a one-to-one function The Inverse Trigonometric Functions
- 5. A function is one-to-one (1–1) if different inputs produce different outputs. That is, f(x) is 1–1 iff for every pair of inputs u v it must be that f(u) f(v). u f(u) v f(v) u = v f(u) = f(v) a one-to-one function u f(u)=f(v)v u = v a non-one-to-one function The Inverse Trigonometric Functions
- 6. A function is one-to-one (1–1) if different inputs produce different outputs. That is, f(x) is 1–1 iff for every pair of inputs u v it must be that f(u) f(v). Example A. a. g(x) = 2x + 1 is 1–1 u f(u) v f(v) u = v f(u) = f(v) a one-to-one function u f(u)=f(v)v u = v a non-one-to-one function The Inverse Trigonometric Functions
- 7. A function is one-to-one (1–1) if different inputs produce different outputs. That is, f(x) is 1–1 iff for every pair of inputs u v it must be that f(u) f(v). u f(u) v f(v) u = v f(u) = f(v) a one-to-one function u f(u)=f(v)v u = v a non-one-to-one function The Inverse Trigonometric Functions Example A. a. g(x) = 2x + 1 is 1–1 because if u v, then 2u 2v so 2u + 1 2v + 1
- 8. A function is one-to-one (1–1) if different inputs produce different outputs. That is, f(x) is 1–1 iff for every pair of inputs u v it must be that f(u) f(v). Example A. a. g(x) = 2x + 1 is 1–1 because if u v, then 2u 2v so 2u + 1 2v + 1 or g(u) g(v). u f(u) v f(v) u = v f(u) = f(v) a one-to-one function u f(u)=f(v)v u = v a non-one-to-one function The Inverse Trigonometric Functions
- 9. A function is one-to-one (1–1) if different inputs produce different outputs. That is, f(x) is 1–1 iff for every pair of inputs u v it must be that f(u) f(v). u f(u) v f(v) u = v f(u) = f(v) a one-to-one function u f(u)=f(v)v u = v a non-one-to-one function The Inverse Trigonometric Functions Example A. a. g(x) = 2x + 1 is 1–1 because if u v, then 2u 2v so 2u + 1 2v + 1 or g(u) g(v). b. f(x) = x2 is not 1–1
- 10. A function is one-to-one (1–1) if different inputs produce different outputs. That is, f(x) is 1–1 iff for every pair of inputs u v it must be that f(u) f(v). u f(u) v f(v) u = v f(u) = f(v) a one-to-one function u f(u)=f(v)v u = v a non-one-to-one function The Inverse Trigonometric Functions Example A. a. g(x) = 2x + 1 is 1–1 because if u v, then 2u 2v so 2u + 1 2v + 1 or g(u) g(v). b. f(x) = x2 is not 1–1 because 3 –3, but f(3) = f(–3) = 9.
- 11. A function is one-to-one (1–1) if different inputs produce different outputs. That is, f(x) is 1–1 iff for every pair of inputs u v it must be that f(u) f(v). u f(u) v f(v) u = v f(u) = f(v) a one-to-one function u f(u)=f(v)v u = v a non-one-to-one function The Inverse Trigonometric Functions Example A. a. g(x) = 2x + 1 is 1–1 because if u v, then 2u 2v so 2u + 1 2v + 1 or g(u) g(v). b. f(x) = x2 is not 1–1 because 3 –3, but f(3) = f(–3) = 9. c. sin(x) is not 1–1 since sin(0) = sin(π) = 0.
- 12. A function is one-to-one (1–1) if different inputs produce different outputs. That is, f(x) is 1–1 iff for every pair of inputs u v it must be that f(u) f(v). u f(u) v f(v) u = v f(u) = f(v) a one-to-one function u f(u)=f(v)v u = v a non-one-to-one function The Inverse Trigonometric Functions Trig-functions are not 1–1. Example A. a. g(x) = 2x + 1 is 1–1 because if u v, then 2u 2v so 2u + 1 2v + 1 or g(u) g(v). b. f(x) = x2 is not 1–1 because 3 –3, but f(3) = f(–3) = 9. c. sin(x) is not 1–1 since sin(0) = sin(π) = 0.
- 13. The inverse relation of a 1–1 function f(x) is itself a function and it’s called the inverse function of f(x). The Inverse Trigonometric Functions
- 14. The inverse relation of a 1–1 function f(x) is itself a function and it’s called the inverse function of f(x). f(u) v f(v) u = v f(u) = f(v) f(x) is a one-to-one function The Inverse Trigonometric Functions f(x) u
- 15. The inverse relation of a 1–1 function f(x) is itself a function and it’s called the inverse function of f(x). f(u) v f(v) u = v f(u) = f(v) f(x) is a one-to-one function The Inverse Trigonometric Functions u f(u) v f(v) u = v f(u) = f(v) f(x) u
- 16. The inverse relation of a 1–1 function f(x) is itself a function and it’s called the inverse function of f(x). It’s denoted as f –1(x). f(u) v f(v) u = v f(u) = f(v) f(x) is a one-to-one function The Inverse Trigonometric Functions u f(u) v f(v) u = v f(u) = f(v) f –1(x) is a well defined function f –1(x)f(x) u
- 17. The inverse relation of a 1–1 function f(x) is itself a function and it’s called the inverse function of f(x). It’s denoted as f –1(x). f(u) v f(v) u = v f(u) = f(v) f(x) is a one-to-one function The Inverse Trigonometric Functions u f(u) v f(v) u = v f(u) = f(v) f –1(x) is a well defined function f –1(x)f(x) Specifically if A denotes the domain and B denotes the range of f(x), then f –1(x) has B as the domain and A as the range u
- 18. The inverse relation of a 1–1 function f(x) is itself a function and it’s called the inverse function of f(x). It’s denoted as f –1(x). f(u) v f(v) u = v f(u) = f(v) f(x) is a one-to-one function The Inverse Trigonometric Functions u f(u) v f(v) u = v f(u) = f(v) f –1(x) is a well defined function f –1(x)f(x) Specifically if A denotes the domain and B denotes the range of f(x), then f –1(x) has B as the domain and A as the range and f –1(x) is 1–1 also. u
- 19. The inverse relation of a 1–1 function f(x) is itself a function and it’s called the inverse function of f(x). It’s denoted as f –1(x). f(u) v f(v) u = v f(u) = f(v) f(x) is a one-to-one function The Inverse Trigonometric Functions u f(u) v f(v) u = v f(u) = f(v) f –1(x) is a well defined function f –1(x)f(x) Specifically if A denotes the domain and B denotes the range of f(x), then f –1(x) has B as the domain and A as the range and f –1(x) is 1–1 also. Furthermore f –1(f(x)) = x u
- 20. The inverse relation of a 1–1 function f(x) is itself a function and it’s called the inverse function of f(x). It’s denoted as f –1(x). f(u) v f(v) u = v f(u) = f(v) f(x) is a one-to-one function The Inverse Trigonometric Functions u f(u) v f(v) u = v f(u) = f(v) f –1(x) is a well defined function f –1(x)f(x) Furthermore f –1(f(x)) = x x A f(x) B f u Specifically if A denotes the domain and B denotes the range of f(x), then f –1(x) has B as the domain and A as the range and f –1(x) is 1–1 also.
- 21. The inverse relation of a 1–1 function f(x) is itself a function and it’s called the inverse function of f(x). It’s denoted as f –1(x). f(u) v f(v) u = v f(u) = f(v) f(x) is a one-to-one function The Inverse Trigonometric Functions u f(u) v f(v) u = v f(u) = f(v) f –1(x) is a well defined function f –1(x)f(x) Furthermore f –1(f(x)) = x x A f(x) B f f –1 f –1(f(x)) = x u Specifically if A denotes the domain and B denotes the range of f(x), then f –1(x) has B as the domain and A as the range and f –1(x) is 1–1 also.
- 22. The inverse relation of a 1–1 function f(x) is itself a function and it’s called the inverse function of f(x). It’s denoted as f –1(x). f(u) v f(v) u = v f(u) = f(v) f(x) is a one-to-one function The Inverse Trigonometric Functions u f(u) v f(v) u = v f(u) = f(v) f –1(x) is a well defined function f –1(x)f(x) x A f(x) B f f –1 f –1(f(x)) = x Furthermore f –1(f(x)) = x and f (f –1(x)) = x u Specifically if A denotes the domain and B denotes the range of f(x), then f –1(x) has B as the domain and A as the range and f –1(x) is 1–1 also.
- 23. The inverse relation of a 1–1 function f(x) is itself a function and it’s called the inverse function of f(x). It’s denoted as f –1(x). f(u) v f(v) u = v f(u) = f(v) f(x) is a one-to-one function The Inverse Trigonometric Functions u f(u) v f(v) u = v f(u) = f(v) f –1(x) is a well defined function f –1(x)f(x) x A f(x) B f f –1 f –1(f(x)) = x Furthermore f –1(f(x)) = x and f (f –1(x)) = x f– 1(x) A B f –1 f –1 (x) x u Specifically if A denotes the domain and B denotes the range of f(x), then f –1(x) has B as the domain and A as the range and f –1(x) is 1–1 also.
- 24. The inverse relation of a 1–1 function f(x) is itself a function and it’s called the inverse function of f(x). It’s denoted as f –1(x). f(u) v f(v) u = v f(u) = f(v) f(x) is a one-to-one function The Inverse Trigonometric Functions u f(u) v f(v) u = v f(u) = f(v) f –1(x) is a well defined function f –1(x)f(x) x A f(x) B f f –1 f –1(f(x)) = x Furthermore f –1(f(x)) = x and f (f –1(x)) = x A B f f –1 f(f –1 (x)) = x xf– 1(x) u Specifically if A denotes the domain and B denotes the range of f(x), then f –1(x) has B as the domain and A as the range and f –1(x) is 1–1 also.
- 25. Some inverse functions may be solved algebraically from f(x). The Inverse Trigonometric Functions
- 26. Some inverse functions may be solved algebraically from f(x). To do this, solve the equation y = f(x) for the variable x in terms of y, the solution is x = f–1(y). The Inverse Trigonometric Functions
- 27. Some inverse functions may be solved algebraically from f(x). To do this, solve the equation y = f(x) for the variable x in terms of y, the solution is x = f–1(y). However, in general it’s not possible to solve algebraically for f–1(x). The Inverse Trigonometric Functions
- 28. Some inverse functions may be solved algebraically from f(x). To do this, solve the equation y = f(x) for the variable x in terms of y, the solution is x = f–1(y). However, in general it’s not possible to solve algebraically for f–1(x). Example B. Solve for f–1(x) given that f(x) = 2x + 1. The Inverse Trigonometric Functions
- 29. Some inverse functions may be solved algebraically from f(x). To do this, solve the equation y = f(x) for the variable x in terms of y, the solution is x = f–1(y). However, in general it’s not possible to solve algebraically for f–1(x). Example B. Solve for f–1(x) given that f(x) = 2x + 1. The Inverse Trigonometric Functions Set y = f(x) = 2x + 1 and solve for x,
- 30. Some inverse functions may be solved algebraically from f(x). To do this, solve the equation y = f(x) for the variable x in terms of y, the solution is x = f–1(y). However, in general it’s not possible to solve algebraically for f–1(x). Example B. Solve for f–1(x) given that f(x) = 2x + 1. The Inverse Trigonometric Functions Set y = f(x) = 2x + 1 and solve for x, we get x = (y – 1)/2 = f–1(y)
- 31. Some inverse functions may be solved algebraically from f(x). To do this, solve the equation y = f(x) for the variable x in terms of y, the solution is x = f–1(y). However, in general it’s not possible to solve algebraically for f–1(x). Example B. Solve for f–1(x) given that f(x) = 2x + 1. The Inverse Trigonometric Functions Set y = f(x) = 2x + 1 and solve for x, we get x = (y – 1)/2 = f–1(y) Switch to the variable x, we’ve f–1(x) = (x – 1)/2.
- 32. Some inverse functions may be solved algebraically from f(x). To do this, solve the equation y = f(x) for the variable x in terms of y, the solution is x = f–1(y). However, in general it’s not possible to solve algebraically for f–1(x). Example B. Solve for f–1(x) given that f(x) = 2x + 1. The Inverse Trigonometric Functions Set y = f(x) = 2x + 1 and solve for x, we get x = (y – 1)/2 = f–1(y) Switch to the variable x, we’ve f–1(x) = (x – 1)/2. If a function f with domain A is not 1–1, we may downsize the domain A so that f is 1–1 in the new domain.
- 33. Some inverse functions may be solved algebraically from f(x). To do this, solve the equation y = f(x) for the variable x in terms of y, the solution is x = f–1(y). However, in general it’s not possible to solve algebraically for f–1(x). Example B. Solve for f–1(x) given that f(x) = 2x + 1. The Inverse Trigonometric Functions Set y = f(x) = 2x + 1 and solve for x, we get x = (y – 1)/2 = f–1(y) Switch to the variable x, we’ve f–1(x) = (x – 1)/2. If a function f with domain A is not 1–1, we may downsize the domain A so that f is 1–1 in the new domain. We then may talk about f and f–1 in relation to this new domain.
- 34. There is no inverse for x2 with the set of all real numbers R as the domain because x2 is not a 1–1 function over R. The Inverse Trigonometric Functions
- 35. There is no inverse for x2 with the set of all real numbers R as the domain because x2 is not a 1–1 function over R. However, if we set the domain to be A = {x ≥ 0} (non–negative numbers) then the function g(x) = x2 is a 1–1 function with the range B = {y ≥ 0}. The Inverse Trigonometric Functions
- 36. There is no inverse for x2 with the set of all real numbers R as the domain because x2 is not a 1–1 function over R. However, if we set the domain to be A = {x ≥ 0} (non–negative numbers) then the function g(x) = x2 is a 1–1 function with the range B = {y ≥ 0}. The Inverse Trigonometric Functions Hence g–1 exists.
- 37. There is no inverse for x2 with the set of all real numbers R as the domain because x2 is not a 1–1 function over R. However, if we set the domain to be A = {x ≥ 0} (non–negative numbers) then the function g(x) = x2 is a 1–1 function with the range B = {y ≥ 0}. The Inverse Trigonometric Functions Hence g–1 exists. To find it, set y = g(x) = x2,
- 38. There is no inverse for x2 with the set of all real numbers R as the domain because x2 is not a 1–1 function over R. However, if we set the domain to be A = {x ≥ 0} (non–negative numbers) then the function g(x) = x2 is a 1–1 function with the range B = {y ≥ 0}. The Inverse Trigonometric Functions Hence g–1 exists. To find it, set y = g(x) = x2, solve for x we’ve x = ±√y.
- 39. There is no inverse for x2 with the set of all real numbers R as the domain because x2 is not a 1–1 function over R. However, if we set the domain to be A = {x ≥ 0} (non–negative numbers) then the function g(x) = x2 is a 1–1 function with the range B = {y ≥ 0}. The Inverse Trigonometric Functions Hence g–1 exists. To find it, set y = g(x) = x2, solve for x we’ve x = ±√y. Since x is non–negative we must have x = √y = g–1(y)
- 40. There is no inverse for x2 with the set of all real numbers R as the domain because x2 is not a 1–1 function over R. However, if we set the domain to be A = {x ≥ 0} (non–negative numbers) then the function g(x) = x2 is a 1–1 function with the range B = {y ≥ 0}. The Inverse Trigonometric Functions Hence g–1 exists. To find it, set y = g(x) = x2, solve for x we’ve x = ±√y. Since x is non–negative we must have x = √y = g–1(y) or that g–1(x) = √x.
- 41. There is no inverse for x2 with the set of all real numbers R as the domain because x2 is not a 1–1 function over R. However, if we set the domain to be A = {x ≥ 0} (non–negative numbers) then the function g(x) = x2 is a 1–1 function with the range B = {y ≥ 0}. The Inverse Trigonometric Functions Hence g–1 exists. To find it, set y = g(x) = x2, solve for x we’ve x = ±√y. Since x is non–negative we must have x = √y = g–1(y) or that g–1(x) = √x. g(x) = x2 g–1(x) = √x Here are their graphs. y = x
- 42. There is no inverse for x2 with the set of all real numbers R as the domain because x2 is not a 1–1 function over R. However, if we set the domain to be A = {x ≥ 0} (non–negative numbers) then the function g(x) = x2 is a 1–1 function with the range B = {y ≥ 0}. The Inverse Trigonometric Functions Hence g–1 exists. To find it, set y = g(x) = x2, solve for x we’ve x = ±√y. Since x is non–negative we must have x = √y = g–1(y) or that g–1(x) = √x. g(x) = x2 g–1(x) = √x Here are their graphs. Note that they are symmetric about the line y = x. y = x
- 43. The Inverse Trigonometric Functions Let f and f–1 be a pair of inverse functions and that f(a) = b
- 44. The Inverse Trigonometric Functions Let f and f–1 be a pair of inverse functions and that f(a) = b so that f–1(b) = a.
- 45. The Inverse Trigonometric Functions y = f(x) y = f–1 (x) (a, b) Let f and f–1 be a pair of inverse functions and that f(a) = b so that f–1(b) = a. Therefore (a, b) is a point on the graph of y = f(x) and that (b, a) is a point on the graph of y = f–1(x). (b, a)
- 46. The Inverse Trigonometric Functions y = f(x) y = f–1 (x) y = x Let f and f–1 be a pair of inverse functions and that f(a) = b so that f–1(b) = a. Therefore (a, b) is a point on the graph of y = f(x) and that (b, a) is a point on the graph of y = f–1(x). The points (a, b) and (b, a) are mirror images with respect to the line y = x. (a, b) (b, a)
- 47. The Inverse Trigonometric Functions y = f(x) y = f–1 (x) y = x Let f and f–1 be a pair of inverse functions and that f(a) = b so that f–1(b) = a. Therefore (a, b) is a point on the graph of y = f(x) and that (b, a) is a point on the graph of y = f–1(x). The points (a, b) and (b, a) are mirror images with respect to the line y = x. (a, b) (b, a) This is true for all points on the graph.
- 48. The Inverse Trigonometric Functions y = f(x) y = f–1 (x) So the graphs of f and f–1 are symmetric diagonally. y = x(a, b) (b, a) Let f and f–1 be a pair of inverse functions and that f(a) = b so that f–1(b) = a. Therefore (a, b) is a point on the graph of y = f(x) and that (b, a) is a point on the graph of y = f–1(x). The points (a, b) and (b, a) are mirror images with respect to the line y = x. This is true for all points on the graph.
- 49. The Inverse Trigonometric Functions Trig–functions are not 1–1 due to their periodicity.
- 50. The Inverse Trigonometric Functions Trig–functions are not 1–1 due to their periodicity. For example cos(0) = cos(2π) = cos(4π) = … = 1.
- 51. The Inverse Trigonometric Functions Trig–functions are not 1–1 due to their periodicity. For example cos(0) = cos(2π) = cos(4π) = … = 1. If we downsize the domain to [0, π], then a = cos(): [0, π] [-1, 1] is a 1–1 function.
- 52. The Inverse Trigonometric Functions Trig–functions are not 1–1 due to their periodicity. For example cos(0) = cos(2π) = cos(4π) = … = 1. If we downsize the domain to [0, π], then a = cos(): [0, π] [-1, 1] is a 1–1 function. Then its inverse exists and it’s denoted as cos–1(a)
- 53. The Inverse Trigonometric Functions Trig–functions are not 1–1 due to their periodicity. For example cos(0) = cos(2π) = cos(4π) = … = 1. If we downsize the domain to [0, π], then a = cos(): [0, π] [-1, 1] is a 1–1 function. Then its inverse exists and it’s denoted as cos–1(a) cos–1(a): [-1, 1] [0, π]
- 54. The Inverse Trigonometric Functions Trig–functions are not 1–1 due to their periodicity. For example cos(0) = cos(2π) = cos(4π) = … = 1. π0 1 –1 a = cos() a If we downsize the domain to [0, π], then a = cos(): [0, π] [-1, 1] is a 1–1 function. Then its inverse exists and it’s denoted as cos–1(a) cos–1(a): [-1, 1] [0, π]
- 55. The Inverse Trigonometric Functions Trig–functions are not 1–1 due to their periodicity. For example cos(0) = cos(2π) = cos(4π) = … = 1. π0 1 –1 a = cos() a If we downsize the domain to [0, π], then a = cos(): [0, π] [-1, 1] is a 1–1 function. Then its inverse exists and it’s denoted as cos–1(a). cos–1(a): [-1, 1] [0, π]
- 56. The Inverse Trigonometric Functions Trig–functions are not 1–1 due to their periodicity. For example cos(0) = cos(2π) = cos(4π) = … = 1. Specifically, given [-1, 1] cos–1(a) = if cos() = a π0 1 –1 a = cos() a If we downsize the domain to [0, π], then a = cos(): [0, π] [-1, 1] is a 1–1 function. Then its inverse exists and it’s denoted as cos–1(a) cos–1(a): [-1, 1] [0, π]
- 57. The Inverse Trigonometric Functions Trig–functions are not 1–1 due to their periodicity. For example cos(0) = cos(2π) = cos(4π) = … = 1. Specifically, given [-1, 1] cos–1(a) = if cos() = a and [0, π]. π0 1 –1 a = cos() a If we downsize the domain to [0, π], then a = cos(): [0, π] [-1, 1] is a 1–1 function. Then its inverse exists and it’s denoted as cos–1(a) cos–1(a): [-1, 1] [0, π]
- 58. The Inverse Trigonometric Functions Trig–functions are not 1–1 due to their periodicity. For example cos(0) = cos(2π) = cos(4π) = … = 1. Specifically, given [-1, 1] cos–1(a) = if cos() = a and [0, π]. Example C. π0 1 –1 a = cos() a If we downsize the domain to [0, π], then a = cos(): [0, π] [-1, 1] is a 1–1 function. Then its inverse exists and it’s denoted as cos–1(a) cos–1(a): [-1, 1] [0, π] a. cos–1(½) = π/3 c. cos–1(cos(5π/3)) = π/3 b. cos–1(–½) = 2π/3
- 59. The Inverse Trigonometric Functions Trig–functions are not 1–1 due to their periodicity. For example cos(0) = cos(2π) = cos(4π) = … = 1. Specifically, given [-1, 1] cos–1(a) = if cos() = a and [0, π]. Example C. π0 1 –1 a = cos() a If we downsize the domain to [0, π], then a = cos(): [0, π] [-1, 1] is a 1–1 function. Then its inverse exists and it’s denoted as cos–1(a) cos–1(a): [-1, 1] [0, π] a. cos–1(½) = π/3 c. cos–1(cos(5π/3)) = π/3 b. cos–1(–½) = 2π/3
- 60. The Inverse Trigonometric Functions Trig–functions are not 1–1 due to their periodicity. For example cos(0) = cos(2π) = cos(4π) = … = 1. Specifically, given [-1, 1] cos–1(a) = if cos() = a and [0, π]. Example C. π0 1 –1 a = cos() a If we downsize the domain to [0, π], then a = cos(): [0, π] [-1, 1] is a 1–1 function. Then its inverse exists and it’s denoted as cos–1(a) cos–1(a): [-1, 1] [0, π] a. cos–1(½) = π/3 c. cos–1(cos(5π/3)) = π/3 b. cos–1(–½) = 2π/3
- 61. The Inverse Trigonometric Functions Trig–functions are not 1–1 due to their periodicity. For example cos(0) = cos(2π) = cos(4π) = … = 1. Specifically, given [-1, 1] cos–1(a) = if cos() = a and [0, π]. Example C. π0 1 –1 a = cos() a If we downsize the domain to [0, π], then a = cos(): [0, π] [-1, 1] is a 1–1 function. Then its inverse exists and it’s denoted as cos–1(a) cos–1(a): [-1, 1] [0, π] a. cos–1(½) = π/3 c. cos–1(cos(5π/3)) = π/3 b. cos–1(–½) = 2π/3
- 62. The Inverse Trigonometric Functions 0 1 –1 y x y = cos(x) If we use x as the input variable and y as the output variable and plot both functions, we have y = cos(x): [0, π] to [–1,1] and its refection across y = x y = cos–1(x): [–1,1] to [0, π] . π
- 63. The Inverse Trigonometric Functions 0 1 –1 y x If we use x as the input variable and y as the output variable and plot both functions, we have y = cos(x): [0, π] to [–1,1] and its refection across y = x y = cos–1(x): [–1,1] to [0, π] . 0 1 y = cos(x) y = x π
- 64. The Inverse Trigonometric Functions 1 –1 y x If we use x as the input variable and y as the output variable and plot both functions, we have y = cos(x): [0, π] to [–1,1] and its refection across y = x y = cos–1(x): [–1,1] to [0, π] . 1 y = cos(x) y = x π –1 0 1 π
- 65. The Inverse Trigonometric Functions π π –1 x y = cos–1(x) y = cos(x) y = x –1 0 1 1 y If we use x as the input variable and y as the output variable and plot both functions, we have y = cos(x): [0, π] to [–1,1] and its refection across y = x y = cos–1(x): [–1,1] to [0, π] .
- 66. The Inverse Trigonometric Functions Using the variables and a and view cos() = a as a function about the unit circle, we have the definition of cos() = a = x–coordinate of the end point (a, b) on the unit circle defined by the arc of length rad. 1 (0, 0)
- 67. The Inverse Trigonometric Functions Using the variables and a and view cos() = a as a function about the unit circle, we have the definition of cos() = a = x–coordinate of the end point (a, b) on the unit circle defined by the arc of length rad. 1 (0, 0) = 0 = π
- 68. The Inverse Trigonometric Functions Using the variables and a and view cos() = a as a function about the unit circle, we have the definition of cos() = a = x–coordinate of the end point (a, b) on the unit circle defined by the arc of length rad. (a, b) 1 (0, 0) = 0 = π
- 69. The Inverse Trigonometric Functions Using the variables and a and view cos() = a as a function about the unit circle, we have the definition of cos() = a = x–coordinate of the end point (a, b) on the unit circle defined by the arc of length rad. = 0 = π (a, b) b 1 (0, 0) a=cos()
- 70. The Inverse Trigonometric Functions Using the variables and a and view cos() = a as a function about the unit circle, we have the definition of cos() = a = x–coordinate of the end point (a, b) on the unit circle defined by the arc of length rad. = 0 = π (a, b) b 1 (0, 0) The cosine inverse cos–1(a) = (–1≤ a ≤ 1) a=cos()
- 71. The Inverse Trigonometric Functions Using the variables and a and view cos() = a as a function about the unit circle, we have the definition of cos() = a = x–coordinate of the end point (a, b) on the unit circle defined by the arc of length rad. (a, b) b 1 (0, 0) = 0 = π a(0, 0) = 0 = π a=cos() The cosine inverse cos–1(a) = (–1≤ a ≤ 1) 1
- 72. The Inverse Trigonometric Functions Using the variables and a and view cos() = a as a function about the unit circle, we have the definition of cos() = a = x–coordinate of the end point (a, b) on the unit circle defined by the arc of length rad. (a, b) b 1 (0, 0) = 0 = π a=cos() The cosine inverse cos–1(a) = (–1≤ a ≤ 1) = length of the arc whose end point has a as the x–coordinate. = 0 = π a(0, 0) 1
- 73. The Inverse Trigonometric Functions Using the variables and a and view cos() = a as a function about the unit circle, we have the definition of cos() = a = x–coordinate of the end point (a, b) on the unit circle defined by the arc of length rad. (a, b) b 1 (0, 0) = 0 = π (a, b) a b 1 (0, 0) = 0 = π a=cos() The cosine inverse cos–1(a) = (–1≤ a ≤ 1) = length of the arc whose end point has a as the x–coordinate.
- 74. The Inverse Trigonometric Functions Using the variables and a and view cos() = a as a function about the unit circle, we have the definition of cos() = a = x–coordinate of the end point (a, b) on the unit circle defined by the arc of length rad. (a, b) b 1 (0, 0) = 0 = π (a, b) a b cos–1(a) = (0, 0) = 0 = π a=cos() The cosine inverse cos–1(a) = (–1≤ a ≤ 1) = length of the arc whose end point has a as the x–coordinate. (0 ≤ ≤ π) 1
- 75. The Inverse Trigonometric Functions Using the variables and a and view cos() = a as a function about the unit circle, we have the definition of cos() = a = x–coordinate of the end point (a, b) on the unit circle defined by the arc of length rad. (a, b) b 1 (0, 0) = 0 = π (a, b) a b (0, 0) = 0 = π a=cos() The cosine inverse cos–1(a) = (–1≤ a ≤ 1) = length of the arc whose end point has a as the x–coordinate. (0 ≤ ≤ π) cos–1(a) = 1 So cos–1(a) is also notated as arccos(a) or acos(a) because cos–1(a) = is the length of a circular arc.
- 76. The Inverse Trigonometric Functions With the variables , a, we’ve the following right triangle represent the relation cos–1() = a .
- 77. The Inverse Trigonometric Functions With the variables , a, we’ve the following right triangle represent the relation cos–1() = a . 1 a = cos–1(b)
- 78. The Inverse Trigonometric Functions With the variables , a, we’ve the following right triangle represent the relation cos–1() = a . Hence the opposite side is √1 – a2. 1 a √1 – a2 = cos–1(b)
- 79. The Inverse Trigonometric Functions With the variables , a, we’ve the following right triangle represent the relation cos–1() = a . Hence the opposite side is √1 – a2. Example D. Draw and find the trig–value. b. cot(cos–1(2a)) a. sin(cos–1(–3/5)) 1 a √1 – a2 = cos–1(b)
- 80. The Inverse Trigonometric Functions With the variables , a, we’ve the following right triangle represent the relation cos–1() = a 1 a . Hence the opposite side is √1 – a2. √1 – a2 Example D. Draw and find the trig–value. b. cot(cos–1(2a)) a. sin(cos–1(–3/5)) –3 5 =cos–1(–3/5) = cos–1(b)
- 81. The Inverse Trigonometric Functions With the variables , a, we’ve the following right triangle represent the relation cos–1() = a 1 a . Hence the opposite side is √1 – a2. √1 – a2 Example D. Draw and find the trig–value. b. cot(cos–1(2a)) a. sin(cos–1(–3/5)) –3 54 =cos–1(–3/5) = cos–1(b)
- 82. The Inverse Trigonometric Functions With the variables , a, we’ve the following right triangle represent the relation cos–1() = a 1 a . Hence the opposite side is √1 – a2. √1 – a2 Example D. Draw and find the trig–value. b. cot(cos–1(2a)) a. sin(cos–1(–3/5)) = 4/5 –3 54 =cos–1(–3/5) = cos–1(b)
- 83. The Inverse Trigonometric Functions With the variables , a, we’ve the following right triangle represent the relation cos–1() = a 1 a . Hence the opposite side is √1 – a2. √1 – a2 Example D. Draw and find the trig–value. b. cot(cos–1(2a)) 1 2a a. sin(cos–1(–3/5)) = 4/5 –3 54 =cos–1(–3/5) = cos–1(b)
- 84. The Inverse Trigonometric Functions With the variables , a, we’ve the following right triangle represent the relation cos–1() = a 1 a . Hence the opposite side is √1 – a2. √1 – a2 Example D. Draw and find the trig–value. b. cot(cos–1(2a)) 1 2a √1 – 4a2 a. sin(cos–1(–3/5)) = 4/5 –3 54 =cos–1(–3/5) = cos–1(b)
- 85. The Inverse Trigonometric Functions With the variables , a, we’ve the following right triangle represent the relation cos–1() = a 1 a . Hence the opposite side is √1 – a2. √1 – a2 Example D. Draw and find the trig–value. b. cot(cos–1(2a)) 1 2a √1 – 4a2 = 2a √1 – 4a2 a. sin(cos–1(–3/5)) = 4/5 –3 54 =cos–1(–3/5) = cos–1(b)
- 86. The Inverse Trigonometric Functions The inverses of the other trig–functions are defined similarly so that each trig–function has its own domain and range.
- 87. The Inverse Trigonometric Functions The inverses of the other trig–functions are defined similarly so that each trig–function has its own domain and range. The sine inverse is defined for –1≤ b ≤ 1.
- 88. The Inverse Trigonometric Functions The inverses of the other trig–functions are defined similarly so that each trig–function has its own domain and range. The sine inverse is defined for –1≤ b ≤ 1. sin–1(b) = ɛ [–π/2, π/2] = length of the arc whose end point has b as the y–coordinate
- 89. The Inverse Trigonometric Functions The inverses of the other trig–functions are defined similarly so that each trig–function has its own domain and range. = π/2 (0, 0) The sine inverse is defined for –1≤ b ≤ 1. sin–1(b) = ɛ [–π/2, π/2] = length of the arc whose end point has b as the y–coordinate = – π/2
- 90. The Inverse Trigonometric Functions The inverses of the other trig–functions are defined similarly so that each trig–function has its own domain and range. = π/2 b (0, 0) The sine inverse is defined for –1≤ b ≤ 1. = – π/2 sin–1(b) = ɛ [–π/2, π/2] = length of the arc whose end point has b as the y–coordinate
- 91. The Inverse Trigonometric Functions The inverses of the other trig–functions are defined similarly so that each trig–function has its own domain and range. = π/2 b (0, 0) The sine inverse is defined for –1≤ b ≤ 1. = – π/2 (a, b)a sin–1(b) = ɛ [–π/2, π/2] = length of the arc whose end point has b as the y–coordinate
- 92. The Inverse Trigonometric Functions The inverses of the other trig–functions are defined similarly so that each trig–function has its own domain and range. = π/2 b (0, 0) sin–1(b) = The sine inverse is defined for –1≤ b ≤ 1. = – π/2 (a, b)a sin–1(b) = ɛ [–π/2, π/2] = length of the arc whose end point has b as the y–coordinate
- 93. The Inverse Trigonometric Functions The inverses of the other trig–functions are defined similarly so that each trig–function has its own domain and range. = π/2 b (0, 0) sin–1(b) = The sine inverse is defined for –1≤ b ≤ 1. = – π/2 (a, b)a The right triangle representing sin–1(b) is shown here. 1 b = sin–1(b) sin–1(b) = ɛ [–π/2, π/2] = length of the arc whose end point has b as the y–coordinate
- 94. The Inverse Trigonometric Functions The inverses of the other trig–functions are defined similarly so that each trig–function has its own domain and range. = π/2 b (0, 0) sin–1(b) = The sine inverse is defined for –1≤ b ≤ 1. = – π/2 (a, b)a The right triangle representing sin–1(b) is shown here. The adjacent of the triangle is √1 – b2 . 1 b = sin–1(b) √1 – b2 sin–1(b) = ɛ [–π/2, π/2] = length of the arc whose end point has b as the y–coordinate
- 95. The Inverse Trigonometric Functions and the inverse of sine y = sin–1(x) : [-1, 1] [-π/2, π/2]. In variables x and y y = sin(x): [-π/2, π/2] [-1, 1].
- 96. The Inverse Trigonometric Functions and the inverse of sine y = sin–1(x) : [-1, 1] [-π/2, π/2]. We get the graph of y = sin–1(x) by reflecting y = sin(x) across y = x. In variables x and y y = sin(x): [-π/2, π/2] [-1, 1].
- 97. The Inverse Trigonometric Functions π/2 y = sin(x) 1 –1 and the inverse of sine y = sin–1(x) : [-1, 1] [-π/2, π/2]. We get the graph of y = sin–1(x) by reflecting y = sin(x) across y = x. –π/2 In variables x and y y = sin(x): [-π/2, π/2] [-1, 1].
- 98. The Inverse Trigonometric Functions π/2 1 –1 y = sin(x) –π/2 and the inverse of sine y = sin–1(x) : [-1, 1] [-π/2, π/2]. We get the graph of y = sin–1(x) by reflecting y = sin(x) across y = x. In variables x and y y = sin(x): [-π/2, π/2] [-1, 1].
- 99. The Inverse Trigonometric Functions –1 –π/2 y = sin–1(x) 1 –1 –π/2 π/2 1–1 π/2 y = x y = sin(x) and the inverse of sine y = sin–1(x) : [-1, 1] [-π/2, π/2]. We get the graph of y = sin–1(x) by reflecting y = sin(x) across y = x. In variables x and y y = sin(x): [-π/2, π/2] [-1, 1].
- 100. The Inverse Trigonometric Functions –π/2 π/2 1–1 y = sin–1(x) and the inverse of sine y = sin–1(x) : [-1, 1] [-π/2, π/2]. We get the graph of y = sin–1(x) by reflecting y = sin(x) across y = x. In variables x and y y = sin(x): [-π/2, π/2] [-1, 1].
- 101. The Inverse Trigonometric Functions Set the domain of tangent to be (–π/2, π/2) so tan() = t: (–π/2, π/2) (–∞ , ∞) is 1–1.
- 102. The Inverse Trigonometric Functions Set the domain of tangent to be (–π/2, π/2) so tan() = t: (–π/2, π/2) (–∞ , ∞) is 1–1. So the tangent inverse exists tan– 1(t) = , –∞< t < ∞
- 103. The Inverse Trigonometric Functions Set the domain of tangent to be (–π/2, π/2) so tan() = t: (–π/2, π/2) (–∞ , ∞) is 1–1. So the tangent inverse exists tan– 1(t) = , –∞< t < ∞ with tan() = t
- 104. The Inverse Trigonometric Functions Set the domain of tangent to be (–π/2, π/2) so tan() = t: (–π/2, π/2) (–∞ , ∞) is 1–1. So the tangent inverse exists tan– 1(t) = , –∞< t < ∞ with tan() = t = b/a where (a, b) is the end point of the arc define by with ɛ (–π/2, π/2) .
- 105. The Inverse Trigonometric Functions = π/2 (0, 0) = –π/2 Set the domain of tangent to be (–π/2, π/2) so tan() = t: (–π/2, π/2) (–∞ , ∞) is 1–1. So the tangent inverse exists tan– 1(t) = , –∞< t < ∞ with tan() = t = b/a where (a, b) is the end point of the arc define by with ɛ (–π/2, π/2) .
- 106. The Inverse Trigonometric Functions = π/2 (0, 0) = –π/2 (a, b), Set the domain of tangent to be (–π/2, π/2) so tan() = t: (–π/2, π/2) (–∞ , ∞) is 1–1. t = b/a So the tangent inverse exists tan– 1(t) = , –∞< t < ∞ with tan() = t = b/a where (a, b) is the end point of the arc define by with ɛ (–π/2, π/2) .
- 107. The Inverse Trigonometric Functions = π/2 b (0, 0) 1 = –π/2 (a, b) a Set the domain of tangent to be (–π/2, π/2) so tan() = t: (–π/2, π/2) (–∞ , ∞) is 1–1. = tan–1(b/a) (a, b), t = b/a So the tangent inverse exists tan– 1(t) = , –∞< t < ∞ with tan() = t = b/a where (a, b) is the end point of the arc define by with ɛ (–π/2, π/2) .
- 108. The Inverse Trigonometric Functions = π/2 b (0, 0) 1 = –π/2 a 1 t = tan–1(t) Set the domain of tangent to be (–π/2, π/2) so tan() = t: (–π/2, π/2) (–∞ , ∞) is 1–1. = tan–1(b/a) (a, b), t = b/a The right triangle representing tan–1(t) is shown here. So the tangent inverse exists tan– 1(t) = , –∞< t < ∞ with tan() = t = b/a where (a, b) is the end point of the arc define by with ɛ (–π/2, π/2) .
- 109. The Inverse Trigonometric Functions = π/2 b (0, 0) 1 = –π/2 a The right triangle representing tan–1(t) is shown here. We note that the hypotenuse is 1 + t2. 1 t = tan–1(t) 1 + t2 Set the domain of tangent to be (–π/2, π/2) so tan() = t: (–π/2, π/2) (–∞ , ∞) is 1–1. = tan–1(b/a) (a, b), t = b/a So the tangent inverse exists tan– 1(t) = , –∞< t < ∞ with tan() = t = b/a where (a, b) is the end point of the arc define by with ɛ (–π/2, π/2) .
- 110. The Inverse Trigonometric Functions –π/2 y = tan(x) Using x and y, we get the graph of y = tan–1(x) by reflecting y = tan(x) across y = x. –π/2
- 111. The Inverse Trigonometric Functions –π/2 y = tan(x) –π/2 Using x and y, we get the graph of y = tan–1(x) by reflecting y = tan(x) across y = x.
- 112. The Inverse Trigonometric Functions –π/2 π/2 y = tan(x) –π/2 – π/2 y = tan–1(x) reflect Using x and y, we get the graph of y = tan–1(x) by reflecting y = tan(x) across y = x.
- 113. The Inverse Trigonometric Functions – π/2 y = tan–1(x) We summarize the sin–1(x), cos–1(x), and sin–1(x) here. π/2 y = cos–1(x) y =sin–1(x) Domain Range Rt– Δ [–1, 1] (–∞, ∞) [0, π] [–π/2, π/2] –1 1 –π/2 π/2 –1 1 π x 1 x 1 x 1 √1+x2 √1–x2 √1–x2 Graph [–1, 1] (–π/2, π/2) x x x
- 114. The Inverse Trigonometric Functions y = cot–1(x) We summarize the sec–1(x), csc–1(x), and cot–1(x) here. π/2 y = sec–1(x) y =csc–1(x) Domain Range Rt– Δ 1 ≤ | x | 1 ≤ | x | 0≤ ≤ π [–π/2, π/2] (–π/2, π/2] x 1 1 x x 1 √1+x2 √x2–1 Graph √x2–1 = π/2 (1, π/2) (–1, –π/2) (1, 0) (–1, π) = 0 (0, π/2) (0,– π/2) = 0 (–∞, ∞) x x x

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