Section 6.1 Inverse Circular Functions Recall

Section 6.1
Inverse Circular Functions
Recall from College Algebra that not every function has an inverse function. In order for a function to have an
inverse function, it must be one-to-one.
Properties of Inverse Functions
The inverse of a function f is denoted by f 1 , read β€œf-inverse”.
Since x and y values switch roles between inverses, if (x, y) is on𝑓(π‘₯), then (y, x) is on 𝑓 (π‘₯ ).
The domain of f is the range of f 1 and the range of f is the domain of f 1 .
The graphs of f and f 1 are reflections of each other across the line 𝑦 = π‘₯.
A function is one-to-one, or has an inverse function, if and only if NO horizontal line crosses the graph
more than once.
Recall the function, 𝑓(π‘₯ ) = π‘₯ .
What is its inverse function? 𝑓
How could we restrict the domain
to make it one-to-one?
What is its domain?
What is its range?
What is its range?
(π‘₯ ) =
Inverse Sine Function
ORIGINAL FUNCTION
INVERSE FUNCTION
π’š = 𝐬𝐒𝐧
π’š = 𝐬𝐒𝐧 𝒙
𝟏
𝒙 OR
π’š = 𝐚𝐫𝐜𝐬𝐒𝐧 𝒙
**y is the angle in the interval
Restricted domain: Input
βˆ’ ,
whose sine value is π‘₯. **
𝝅 𝝅
Angles from βˆ’ 𝟐 , 𝟐
Domain: Input
Sine values from[βˆ’1,1]
Range: Output
Range: Output
Sine values from[βˆ’1,1]
Angles from βˆ’ ,
Example 2: Find the exact value of y in each equation.
a)
𝑦 = arcsin
√
b) 𝑦 = sin
βˆ’
Inverse Cosine Function
ORIGINAL FUNCTION
π’š = 𝐜𝐨𝐬 𝒙
INVERSE FUNCTION
π’š = 𝐜𝐨𝐬
𝟏
𝒙 OR
π’š = 𝐚𝐫𝐜𝐜𝐨𝐬 𝒙
**y is angle in the interval [𝟎, 𝝅]whose cosine value is π‘₯. **
Restricted domain: Input
Domain: Input
Angles from [𝟎, 𝝅]
Cosine values
from[βˆ’1,1]
Range: Output
Coine values from[βˆ’1,1]
Range: Output
Angles from [𝟎, 𝝅]
Example 2: Find the exact value of y in each equation.
a)
𝑦 = arccos
√
b) 𝑦 = cos
βˆ’
Inverse Tangent Function
ORIGINAL FUNCTION
π’š = 𝐭𝐚𝐧 𝒙
INVERSE FUNCTION
π’š = 𝐭𝐚𝐧
𝟏
𝒙 OR
π’š = 𝐚𝐫𝐜𝐭𝐚𝐧 𝒙
**y is the angle in the interval
Restricted domain: Input
Angles from βˆ’ ,
βˆ’ ,
whose tangent value is π‘₯. **
Domain: Input
Tangent values from(βˆ’βˆž, ∞)
Range: Output
Tangent values from
(βˆ’βˆž, ∞)
Range: Output
Angles from βˆ’ ,
Example 3: Find the exact value of y in each equation.
a)
𝑦 = arctan 1
b) 𝑦 = tan
βˆ’βˆš3
Inverse Functions for Cotangent, Secant, and Cosecant
π‘₯ can be evaluated by
cot
To evaluate these inverse functions, we can use the reciprocal relationship.
sec
π‘₯
𝐜𝐨𝐬
can be evaluated by
𝟏𝟏
𝒙
csc
𝐭𝐚𝐧
𝟏𝟏
𝒙
if x > 0
𝐭𝐚𝐧
𝟏𝟏
+
𝒙
πŸπŸ–πŸŽ° if x < 0
π‘₯ can be evaluated by
𝐬𝐒𝐧
𝟏𝟏
𝒙
*NOTE: cot
0=
Example 4: For each expression, find the exact measure, in degrees, if it exists. Do not use a calculator.
a)
arccsc βˆ’
√
b) arcsec(βˆ’
√
)
c) cot
βˆ’
√
Example 5: Use a calculator to give each value in decimal degrees. Round to the nearest 2 decimal places.
a) πœƒ = arccos
(βˆ’0.13348122)
b) πœƒ = csc
1.9422833
Example 6: Use a calculator to give each real number value. (Radians) Round to the nearest 6 decimal places.
a) 𝑦 = sin (βˆ’0.13348122)
b) 𝑦 = arcsec(βˆ’1.2671684)
Summary of the domain and ranges of the Inverse Functions
Inverse Function
Domain
RANGE
Interval in Degrees
Quadrants
[βˆ’πŸ—πŸŽ°, πŸ—πŸŽ°]
I and IV
𝝅
𝝅
, 𝟎) βˆͺ (𝟎,
𝟐
𝟐
[βˆ’πŸ—πŸŽ°, 𝟎°) βˆͺ (𝟎°,πŸ—πŸŽ°]
I and IV
𝝅 𝝅
,
𝟐 𝟐
(βˆ’πŸ—πŸŽ°, πŸ—πŸŽ°)
I and IV
Interval in Radians
𝝅 𝝅
βˆ’ ,
𝟐 𝟐
𝑦 = sin
π‘₯
[βˆ’1,1]
𝑦 = csc
π‘₯
(βˆ’βˆž, βˆ’1] βˆͺ [1, ∞)
𝑦 = tan
π‘₯
(βˆ’βˆž, ∞)
βˆ’
𝑦 = cot
π‘₯
(βˆ’βˆž, ∞)
(𝟎, 𝝅)
(𝟎°, πŸπŸ–πŸŽ°)
I and II
𝑦 = cos
π‘₯
[βˆ’1,1]
[𝟎, 𝝅]
[𝟎°, πŸπŸ–πŸŽ°]
I and II
𝑦 = sec
π‘₯
(βˆ’βˆž, βˆ’1] βˆͺ [1, ∞)
[𝟎°, πŸ—πŸŽ°) βˆͺ (πŸ—πŸŽ°, πŸπŸ–πŸŽ°]
I and II
βˆ’
𝟎,
𝝅
𝝅
βˆͺ
,𝝅
𝟐
𝟐