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inverse_trig_functions2

# inverse_trig_functions2 - Inverse Trig Functions By Richard...

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Inverse Trig Functions By Richard Gill Supported in Part by a Grant from VCCS LearningWare

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Let us begin with a simple question: x x f x x f = = - ) ( ) ( 1 2 What is the first pair of inverse functions that pop into YOUR mind? This may not be your pair but this is a famous pair. But something is not quite right with this pair. Do you know what is wrong? Congratulations if you guessed that the top function does not really have an inverse because it is not 1-1 and therefore, the graph will not pass the horizontal line test.
Consider the graph of . 2 x y = -2.0 2.0 2.0 4.0 6.0 8.0 10.0 x y Note the two points on the graph and also on the line y=4. f(2) = 4 and f(-2) = 4 so what is an inverse function supposed to do with 4? ? 2 ) 4 ( 2 ) 4 ( 1 1 - = = - - f or f By definition, a function cannot generate two different outputs for the same input, so the sad truth is that this function, as is, does not have an inverse.

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So how is it that we arrange for this function to have an inverse? We consider only one half of the graph: x > 0. The graph now passes the horizontal line test and we do have an inverse: x x f x for x x f = = - ) ( 0 ) ( 1 2 Note how each graph reflects across the line y = x onto its inverse. x y = 2.0 4.0 6.0 8.0 x 4 y=x 2 x y = 2
A similar restriction on the domain is necessary to create an inverse function for each trig function. Consider the sine function. You can see right away that the sine function does not pass the horizontal line test. But we can come up with a valid inverse function if we restrict the domain as we did with the previous function. How would YOU restrict the domain? -2π/3 -π/3 π/3 2π/3 π 4π/3 5π/3 -1.0 1.0 x y y = sin(x) y = 1/2

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Take a look at the piece of the graph in the red frame. -2π/3 -π/3 π/3 2π/3 π 4π/3 5π/3 -1.0 1.0 x y We are going to build the inverse function from this section of the sine curve because: This section picks up all the outputs of the sine from –1 to 1. This section includes the origin. Quadrant I angles generate the positive ratios and negative angles in Quadrant IV generate the negative ratios. Lets zoom in and look at some key points in this section.
-π/2 -π/3 -π/6 π/6 π/3 π/2 -1.0 1.0 x y y = sin(x) 1 2 2 3 3 2 2 4 2 1 6 0 0 2 1 6 2 2 4 2 3 3 1 2 ) ( π π π π π π π π - - - - - - - - x f x I have plotted the special angles on the curve and the table.

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2 1 3 2 3 4 2 2 6 2 1 0 0 6 2 1 4 2 2 3 2 3 2 1 ) ( sin 1 π π π π π π π π - - - - - - - - - x x The new table generates the graph of the inverse. 1 2 2 3 3 2 2 4 2 1 6 0 0 2 1 6 2 2 4 2 3 3 1 2 ) sin( π π π π π π π π - - - - - - - - x x To get a good look at the graph of the inverse function, we will “turn the tables” on the sine function. The domain of the chosen section of the sine is So the range of the arcsin is - 2 , 2 π π - 2 , 2 π π The range of the chosen section of the sine is [-1 ,1] so the domain of the arcsin is [-1, 1].
Note how each point on the original graph gets “reflected” onto the graph of the inverse.

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