Download sin sin tan tan cos cos

MA 15400
Lesson 16
Section 7.6
The Inverse Trigonometric Functions
Find θ in the interval [0, 2π) for each statement.
1
3
tan θ =
sin θ =
3
2
π 7π
π 2π
θ= ,
θ= ,
6 6
3 3
1
cosθ =
θ=
2
π 7π
4
,
4
While this is true for equations with the directions ‘Find all solutions of the equation in the
interval [0, 2π)’, today we are going to define the inverse trigonometric functions.
INVERSE TRIGONOMETRIC FUNCTIONS:
Remember that a function must pass a vertical line test. And, for each x, there is only one y. For
a trigonometric function, each angle yields only one trig value (a ratio). For an inverse
trigonometric function, each trig value (ratio) would have to yield only one angle. However, we
1
can correspond to more than one angle, for
can see from above, a trig value, such as
3
π
7π
example both
and
. Your calculator is ‘set up’ to return only one angle for each trig
6
6
value. Remember that the inverse sine key or inverse tangent key only return a value in
quadrants I or IV. The inverse cosine key only returns an angle in quadrants I or II.
Notation for Inverse Trigonometric Functions:
y = arcsin(x) = sin-1(x) = θ y =arccos(x) = cos-1 (x)= θ
y = arctan(x) = tan-1(x) = θ
 1 
 1  π
arctan
 = tan −1 
 =
 3
 3 6
 3
 3 π
 = sin −1 

arcsin

 2 = 3
 2 


 1 
 1  π
arccos
 = cos −1 
=
 2
 2 4
Why only one answer? Remember you calculators only returned values in certain quadrants when we
were working with inverce functions? arcsin is only defined from −
quadrants I and IV. arctan is also only defined from −
π
to
π
2
to
π
2
inclusively, which are
π
, which are quadrants I and IV.
2
2
However, arccos is defined from 0 to π inclusively, which are quadrants I and II.
This is reinforced when you use the inv tan, inv cos, or inv sin keys on you calculator.
-1
tan-1
sin-1
tan-1
sin
cos-1
cos-1
Notice: For each function, one quadrant has positive values and the other quadrant
has negative values.
1
Remember: An
inverse trig function
returns an angle.
Given a trig value as
x (a ratio), it returns
a y that is an angle!
MA 15400
Lesson 16
Section 7.6
The Inverse Trigonometric Functions
Summary:
 π π
arcsin is defined in a range of  − ,  and domain of [−1,1]
 2 2
π
π
 π π
arctan is defined in a range of  − ,  (Remember, at − and , the tangent is undefined) and
2
2
 2 2
domain of all real numbers
arccos is defined in a range of [ 0, π ] and domain of [−1,1]
Find the exact value of the expression whenever it is defined.
 1
sin −1  − 
 2
 2

cos −1 

2


tan −1 (− 1)
1
arcsin 
2

3

arccos −

 2 
arctan 3
( )
For the following problems: Always work from the inside out!!!

 2 

sin arcsin
cos[cos −1 (1)]
tan tan −1 (5)

 2 

[
]
You will notice when the trig function is on the
‘outside’, the two functions ‘eliminate’ each
other and the result is the trig value. This may
not happen when the inverse function is on the
‘outside’. Therefore, always work from the
inside out.
2
MA 15400
Lesson 16
Section 7.6
The Inverse Trigonometric Functions
 5π 
arcsin sin

4 

5π 

arccos cos 
4 

5π 

arctan tan

4 

Notice: In the above 3 examples, the argument in each was 5π . However, none of the answers were 5π . It is outside of
4
4
the range. Also, notice that the 3 answers are not the same.


3 

sin cos −1  −


 2 

 1 

cos  tan −1 
 3 


 1 
tan sin −1  − 
 2 

Find the exact value of the expression whenever it is defined.

 5 
sin  2 arccos  −  
 13  

= sin [ 2θ ]
= 2 sin θ cos θ
 5
θ = arccos  − 
 13  T
5
∴ cos θ = −
13
The angle θ is in
Q II.

 4 
cos 2 sin −1  
 5 

3
4
 
4
∴ sin θ =
5
The angle θ is in
QI
θ = sin −1  
5
MA 15400
Lesson 16
Section 7.6
The Inverse Trigonometric Functions

 9 
tan 2 tan −1  − 
 40 

Write the expression as an algebraic expression in x for x > 0.
x
sin θ =
−1
cos ( sin x )
1
sin ( 2 arccos x )
= cos θ
= sin 2θ
1 − x2
=
1
1
cos θ =
x
1
1
x
θ
b
θ
x
x 2 + b 2 = 12
b2 = 1 − x2
= 1 − x2
b = 1 − x2
Proof (using a calculator)
cos(sin −1 x) = 1 − x 2
Let x = 0.26
cos(sin −1 (0.26) = 1 − (0.26) 2
cos(0.263022203) = 1 − 0.0676
0.9656086 = 0.9324
0.9656086 = 0.9656086
4
MA 15400
Lesson 16
Section 7.6
The Inverse Trigonometric Functions
Most function graphs
continue infinitely.
This graph ‘fits’ in a
rectangle.
Graph:
π
y = sin −1 x
2
Trig Value Angle
-1
π
−
2
π
3
−
−
3
2
1
π
−
−
4
2
1
π
−
−
2
6
-1
0
0
1
π
2
6
1
π
4
2
3
2
1
π
3
π
4
π
6
1
−
−
π
π
6
π
4
−
3
π
3
π
2
y = tan −1 x
−
π
2
π
2
π
3
π
4
π
6
2
-2
−
−
6
π
4
−
−
π
π
3
π
2
5
MA 15400
Lesson 16
Section 7.6
The Inverse Trigonometric Functions
y = cos −1 x
π
5π
6
3π
4
2π
3
π
2
π
3
π
4
π
6
-1
1
1
y = cos −1 x
2
π
5π
6
3π
4
2π
3
π
2
π
3
π
4
π
6
-1
6
1
MA 15400
Lesson 16
Section 7.6
The Inverse Trigonometric Functions
y = cos −1 (2 x)
π
5π
6
3π
4
2π
3
π
2
π
3
π
4
π
6
-1
1
7