crms calculus 2010 february 8, 2010_a

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Review of inverse trig functions

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Page 1: CRMS Calculus 2010 February 8, 2010_A

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www.blakeyart.com/lookingback.jpg

inverse trig functions

Looking back on 

Page 2: CRMS Calculus 2010 February 8, 2010_A

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Calculus Group Members ________________________Section 4.5B: Derivatives of Inverse Trigonometric Functions Date ______________

Review:   Inverse Trigonometric Functions

1. Label the graph of each trig function, then state its domain and range.

2. Since the trig functions are all periodic graphs, none of them pass the __________________ _________ test.

    Therefore none of these are _________________ functions or  ___ to ___ functions,. and so do not have    inverses which are functions.    

Rather these are called inverse trig _________________.

Sketch (dashed line type) the graph of the inverse relation for each trigonometric function.

[On TI­83/84, with function in Y1: à Draw à 8:DrawInv]

3. In order to create inverses which are functions, we must restrict the _____________ of each of the functions    so that they are 1­1. 

    This corresponds to considering only a particular __________ of the Unit Circle. (See figure)

The notation for inverse the trig function is y = ______________.

The portion of the graph used for the inverse function is called the ______________  _________.

On the sketch of the inverse trig relation, darken (solid line type) the principle branch of the inverse trigonometric function, then state its domain and range.

Page 3: CRMS Calculus 2010 February 8, 2010_A

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[­1, 1][­1, 1]

[­1, 1][­1, 1]

Page 4: CRMS Calculus 2010 February 8, 2010_A

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Domain: _______________________

Range:  ________________________

Domain: _______________________

Range:  ________________________

except

except

[to graph on TI: cos­1(1/x)]

Page 5: CRMS Calculus 2010 February 8, 2010_A

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Domain: _______________________

Range:  ________________________

Domain: _______________________

Range:  ________________________

[to graph on TI: π/2 ­ tan­1(x)]

[to graph on TI: sin­1(1/x)]