CH 4.7

Practice Test Questions? We will only spend the first half an hour on the Practice Test so make sure you start reviewing right away

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

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.

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)?

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

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

Use the y = sin(x) graph (with the restricted domain) on your handout to explore the relationship between sin(x) and arcsin(x)

Y = arcsin(x)

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

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

Use the y = cos (x) graph (with the restricted domain) on your handout to explore the relationship between cos(x) and arccos(x)

In Summary…

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) = - 3 ?”

You Try! #1-9 odd and #33-39 odd

Exploring Inverse Functions more…

EX 2 If possible, find the exact value: a) arcsin(sin 3п 2 )

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]

You Try!

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 [ , ]?”

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

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