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INSTRUCTORS Raffi Hovasapian John Zhu
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For more information, please see full course syllabus of Calculus AB
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Lecture Comments (1)

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Post by Phil Beauchamp on July 11, 2014

Should there be a negative one in the second inverse sine example that is undefined or is it positive one in both examples?  Should it look like the following?

sin-1(>1) is undefined ?  and
sin-1(<-1) is undefined ?

Inverse Trigonometric Functions

  • Get familiar with shapes: graph with calculator
  • Meaning of inverse: if , then

Inverse Trigonometric Functions

Given f([(π)/6]) = [1/2] with f(x) = sinx and g(x) = sin−1 x, find g([1/2])
  • An inverse function will "reverse" the action done by the original function. Kind of like working backwards.
  • g(x) is the inverse of f(x)
g([1/2]) = [(π)/6]
Find the domain of f(x) = sin−1x
  • The only valid inputs for x are given by the range of its inverse, sinx, [-1, 1]
Domain [-1, 1]
Find sin−1([(√2)/2])
  • sin([(π)/4]) = [(√2)/2]
sin−1([(√2)/2]) = [(π)/4]
There are technically many inputs for sin(x) that give [(√2)/2], but remember it must remain in the range of sin−1 x, which is [−[(π)/2], [(π)/2]].
Find cos−1(1)
  • Range of cos−1 x is [0,π].
  • cos(0) = 1
cos−1(1) = 0
Find sin(cos−1[(√2)/2])
  • cos([(π)/4]) = [(√2)/2]
  • cos−1([(√2)/2]) = [(π)/4]
  • sin(cos−1[(√2)/2]) = sin([(π)/4])
= [(√2)/2]
Find sin−1[1/2]
sin([(π)/6]) = [1/2]
sin−1([1/2]) = [(π)/6]
Find tan(cos−1[(√2)/2])
cos−1[(√2)/2] = [(π)/4]
tan(cos−1[(√2)/2]) = tan[(π)/4]
= 1
Given f(x) = sin2 x and g(x) = sin−1 (sin3x), evaluate f(g(x))
  • g(x) = sin−1 (sin3x)
  • = 3x
f(g(x)) = sin23x
SPECIAL NOTE: By convention sin−1 x is the inverse of sinx. (sinx)−1 is cscx.
Find at least one solution for θ given sin(θ) = 0
  • Take the inverse sin of both sides
  • sin(θ) = 0
  • sin−1(sin(θ)) = sin−1(0)
θ = 0
Graph f(x) = 2 sin−1(3x + 1)
  • We know the input for inverse sin must be within [-1, 1]. Therefore, −1 ≤ 3x + 1 ≤ 1. So [(−2)/3] ≤ x ≤ 0. Domain [[(−2)/3], 0]. Plugging in some values...
  • f([(−2)/3]) = 2 sin−1(−1) = 2 [(− π)/2] = −π
  • f([(−1)/3]) = 2 sin−1(0) = 0
  • f(0) = 2 sin−11 = 2 [(π)/2] = π
  • Increasing function with Range [−π, π]

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Inverse Trigonometric Functions

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Types of Functions: Trigonometric- Inverse Trig Functions 0:07
  • Example: Inverse SIN of X 0:45
  • Example: Inverse Function 2:30
  • Example: Inverse TAN of X 4:42