4.7 Inverse Trig Functions

Does the Sine function have an inverse?

1 -1

What could we restrict the domain to so that the sine function does have an inverse?

   2 ,  2   1 -1

• • • • • Function is increasing Takes on full range of values Function is 1-1 Domain:   1, 1  Range:  2 ,  2  

Evaluate: arcSin 2 3 • Asking the sine of what angle is 2 3

Find the following: a) ArcSin 2 2 b) Sin 1 (  1 2 ) c) ArcSin  2 3

Inverse Cosine Function • What can we restrict the domain of the cosine curve to so that it is 1-1?

 0 ,   1 -1

• • • • • Function is increasing Takes on full range of values Function is 1-1 Domain:   1 , 1  Range:  2 ,  2  

Evaluate: ArcCos (-1) • The Cosine of what angle is -1?

Evaluate the following: a) Cos 1 ( b) ArcCos 2 3 ) ( 1 2 ) c) Cos 1 ( 2 2 )

ArcTan (x) • Similar to the ArcSin (x) • Domain of Tan Function: • Range of Tan Function:

arcCos (0.28) • Is the value 0.28 on either triangle or curve?

• Use your calculator: – Cos -1 (0.28)

Determine the missing Coordinate

Determine the missing Coordinate

Use an inverse trig function to write θ as a function of x.

2x θ x + 3

Find the exact value of the expression.

Sin [ ArcCos     2 3    ]

4.7 Inverse Trig Functions

So far we have: 1) Restricted the domain of trig functions to find their inverse 2) Evaluated inverse trig functions for exact values 3) Found missing coordinates on the graphs of inverses 4) Found the exact values of compositions

Composition of Functions 1) Evaluate innermost function first 2) Substitute in that value 3) Evaluate outermost function Remember t hat f (f -1 (x) )  x and f -1 (f (x) )  x As long as x is in the domain of the necessasry function

Sin (arcCos 1 2 ) Evaluate the innermost function first: arcCos ½ = Substitute that value in original problem  Sin 3

Cos 1    Sin 7  6   

Tan    Cos 1 5 13    How do we evaluate this?

Let θ equal what is in parentheses   Cos 1 5 13 

Cos

  5 13

Cos

  5 13 13 θ 5 12

Tan    Cos 1 5 13     How do we evaluate this?

Tan  Let θ equal what is in parentheses Use the triangle to answer the question Tan   5 12 13 θ 5 12

Csc    Tan 1 15 8   

Sin  Sin -1 0.2

 What is different about this problem?

Is 0.2 in the domain of the arcSin?

Then Sin  Sin -1 0.2

  0 .

2

Sin 1    Sin 4  3    What is different about this problem?

Is 4  3 in the domain of the Sin function?

Since it is not, we must evaluate Sin 4  3

Graph of the ArcSin

Y

  2   3   6  0 6  3  2

X = Sin Y

 1  1 2  3 2 1 2 0 3 2 1

Graph of the ArcSin

Graph of ArcCos

Y

0  6  3  2  2 3 5   6

X = Sin Y

1 3 2 1 2 0  1 2  3  1 2

Graph of the ArcCos