WWe know that for a function to have an inverse function, it must be one-to-one (it must pass the Horizontal Line Test).
FFrom looking at a sine wave, it is obvious that it does not pass the Horizontal Line Test.
The Unit Circle in found in section 4.7. We will use: ◦ Radians ◦ Exact answers (mostly) ◦ Quick board review of Unit Circle, quadrants on the wave, & converting to radian measure
IIn order to pass the Horizontal Line Test (so that sin x has an inverse that is a function), we must restrict the domain. WWe restrict it to
QQuadrant IV is QQuadrant I is AAnswers must be in one of those two quadrants or the answer doesn’t exist.
HHow do we draw inverse functions? SSwitch the x’s and y’s! Switching the x’s and y’s also means switching the axis!
DDomain/range of restricted sine wave? DDomain/range of inverse?
yy = arcsin x or y = sin -1 x BBoth mean the same thing. They mean that you’re looking for the angle where sin y = x.
FFind the exact value of: AArcsin ½ ◦T◦This means at what angle is the sin = ½ ? ◦π◦π/6 ◦(◦(5π/6 has the same answer, but falls in QIII, so it is not correct)
When looking for an inverse answer on the calculator, use the 2 nd key first, then hit sin, cos, or tan. When looking for an angle always hit the 2 nd key first. Last example: ◦ Degree mode, 2 nd, sin,.5 = 30. ◦ Radian mode: 2 nd, sin,.5 =.524 (which is pi/6)
FFind the value of: SSin -1 2 ◦T◦This means at what angle is the sin = 2 ? ◦W◦What does your calculator read? Why? ◦2◦2 falls outside the domain of an inverse sine wave
WWe must restrict the domain NNow the inverse
Graphing Utility: Graphs of Inverse Functions Graphing Utility: Graph the following inverse functions. a. y = arcsin x b. y = arccos x c. y = arctan x –1.5 1.5 –– –1.5 1.5 22 –– –3 3 –– Set calculator to radian mode.
Graphing Utility: Inverse Functions Graphing Utility: Approximate the value of each expression. a. cos – 1 0.75b. arcsin 0.19 c. arctan 1.32d. arcsin 2.5 Set calculator to radian mode.
Previously learned notation: ◦ f o g(x)g o f(x)
FFind tan(arctan(-5)) ◦-◦-5 (the tangent and its inverse cancel each other out!) FFind arcsin(sin ) ◦T◦The domain of the sine function is. Since is outside that domain, we’ll just say that the answer is: is outside the domain (unless you remember coterminal angles and can tell me the actual answer is FFind ◦O◦Outside the domain
Find the exact value of Draw the graph using only the info inside the parentheses. (Easy way—completely ignore the fact that you have inverses!)