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CH 4.7
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Practice Test Questions?
We will only spend the first half an hour on the Practice Test so make sure you start reviewing right away
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Definition of an Inverse Function
An inverse function is a function that "reverses" another function. In order for a function to have an Inverse function, each x must have A unique y (no y-values repeat) A quick way to check for this is with the horizontal line test
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Formal Notation of the inverse trig functions
You might see the inverse sine function as y=arcsin(x) or y = 𝑠𝑖𝑛 −1 (x) Similarly, you might see the inverse cosine function denoted y=arccos(x) or y = 𝑐𝑜𝑠 −1 (x) This pattern is the same for all inverse trig functions. *the notation arcsin comes from the association of a central angle with its intercepted arc length on the unit circle. *the notation 𝑠𝑖𝑛 −1 is consistent with the regular inverse function notation.
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Warm Up 4.7 Recall that for a function to have an inverse, each x must have a unique y. The horizontal line test is a quick way to determine if a function would have an inverse function. Does the sin(x) function pass the horizontal line test? Does the cos(x)?
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If we restrict the domain of the sine function, we can get a unique inverse function.
On the interval [ - П 2 , П 2 ], the function y=sin(x) is one-to-one and increasing. The range takes on all the y values between -1 and 1
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What is the graph of arcsin(x)?
y = sin(x) y = arcsin(x) Switch the domain and range to find the inverse of any function x Y - п 2 - п 4 - п 6 п 6 п 4 п 2 x y
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Use the y = sin(x) graph (with the restricted domain) on your handout to explore the relationship between sin(x) and arcsin(x)
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Y = arcsin(x)
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Restrict the domain of y=cos(x) to get an inverse function
Similar to the way we found the inverse sine function, if we restrict the domain of the cosine function, we can find a unique inverse function. On the interval [ 0, п ], the function y= cos(x) is one-to-one and decreasing. The range takes on all the y values between -1 and 1
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What is the graph of arccos(x)?
y = cos (x) y = arccos(x) Switch the domain and range to find the inverse of any function x Y п 6 п 3 п 2 2п 3 5п 6 п x y
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Use the y = cos (x) graph (with the restricted domain) on your handout to explore the relationship between cos(x) and arccos(x)
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In Summary…
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Examples arcsin(1/2) 3) Arcsin(0) 2) Arccos(1/2) 4) Arctan(- 3 )
“at what angle, in our range of [− П 2 , П 2 ] “at what angle, in our range of [− П 2 , П 2 ] Is the sin(x) = ½ ?” Is the sin(x) = 0 ?” 2) Arccos(1/2) 4) Arctan(- 3 ) “at what angle, in our range of [0,П] “at what angle, in our range of [− П 2 , П 2 ], Is the cos(x) = ½ ?” is the tan(x) = ?”
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You Try! #1-9 odd and #33-39 odd
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Exploring Inverse Functions more…
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EX 2 If possible, find the exact value: a) arcsin(sin 3п 2 )
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EX 2 CONT b) tan[arctan(-5)] c) arcsin(sin 5п 3 ) d) cos(arccos п)
Q1: is -5 in the domain of arctan? -5 -pi/3 Not defined Q1: what is sine of 5п 3 ? Q2: is that value in the domain of arcsin? Recall that the domain of arccos is [-1, 1]
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You Try!
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EX 3 Find the exact value of tan(arccos 2 3 ) b) cos[arcsin( −3 5 )]
Step 1: Make substitutions Step 1: Make substitutions “When is cos(u) = 2/3 in our interval [0, pi]?” “When is sin(u) = -3/5 in our interval [ , ]?”
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You Try with your group! Find the exact value of tan(arccos 2 3 )
Step 1: Make substitutions “When is cos(u) = 2/3 in our interval [0, pi]?” Let u = arccos( 2 3 ) so then cos(u) = 2 3 Step 2: Draw a triangle Because cos(u) is positive u is in the first quadrant Solve for missing sides Step 3: Solve
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You Try Cont. Find the exact value of b) cos[arcsin( −3 5 )]
Step 1: Make substitutions Let u = arcsin( −3 5 ) so then sin(u) = −3 5 Step 2: Draw a triangle Because sin(u) is positive u is in the first quadrant Solve for missing sides Step 3: Solve
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