The Inverse Trigonometric Functions · 2020. 4. 22. · The Inverse Trigonometric Functions By...
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Chapter 20 Additional Topics in Trigonometry ● Section 20.6 The Inverse Trigonometric Functions
Transcript of The Inverse Trigonometric Functions · 2020. 4. 22. · The Inverse Trigonometric Functions By...
Chapter 20 Additional Topics in Trigonometry
● Section 20.6 The Inverse Trigonometric Functions
The Inverse Trigonometric Functions
By solving the equation y = sin x for the independent variable x,
we will get:
x = sin -1 y, −π2
≤x≤π2
The Inverse Trigonometric Functions
By solving the equation y = sin x for the independent variable x,
we will get:
x = sin -1 y, −π2
≤x≤π2
-1
-0.5
0
0.5
1
-3π/2 -π -π/2 0 π/2 π 3π/2 2π-2π
x
ysin(x)
-360° -270° -180° -90° 90° 180° 270° 360°
The Inverse Trigonometric Functions
By solving the equation y = sin x for the independent variable x,
we will get:
x = sin -1 y, −π2
≤x≤π2
-1
-0.5
0
0.5
1
-3π/2 -π -π/2 0 π/2 π 3π/2 2π-2π
x
ysin(x)
-360° -270° -180° -90° 90° 180° 270° 360°
Recall that:a function is invertible if it is one-to-one, i.e. there is only one value y for each value x.
The Inverse Trigonometric Functions
−π2
≤sin−1 x≤π2
0≤cos−1 x≤π
-1
-0.5
0
0.5
1
-3π/2 -π -π/2 0 π/2 π 3π/2 2π-2π
x
ycos(x) sin(x)
-360° -270° -180° -90° 90° 180° 270° 360°
y
x
sin−1 x
cos−1 x
The Inverse Trigonometric Functions
−π2
< tan−1 y≤π2
0<cot−1 y≤π
π2
−π2
−π 0 ππ2
0 −π2
−π π
tan x cot x
The Inverse Trigonometric Functions
−π2
< tan−1 x≤π2 0<cot−1 x≤π
−π2
0
π
π2
The Inverse Trigonometric Functions
−π2
≤csc−1 y≤π2
0≤sec−1 y≤π
csc−1 y≠0 sec−1 y≠π2
π2
−π2 0−π π
The Inverse Trigonometric Functions
−π2
≤csc−1 x≤π2
0≤sec−1 x≤π
csc−1 x≠0 sec−1 x≠π2
π2
π2
0
−π2
π
The Inverse Trigonometric Functions
−π2
≤sin−1 x≤π2
0≤cos−1 x≤π
−π2
≤csc−1 x≤π2
csc−1 x≠0
0≤sec−1 x≤πsec−1 x≠π
2
−π2
< tan−1 x≤π2
0<cot−1 x≤π
Example 1: Find
(a)
(b)
(c)
sin−1 √32
sec−1√2
sin−1−√32
Values of Trigonometric Functions
The Inverse Trigonometric Functions
−π2
≤sin−1 x≤π2
0≤cos−1 x≤π
−π2
≤csc−1 x≤π2
csc−1 x≠0
0≤sec−1 x≤πsec−1 x≠π
2
−π2
< tan−1 x≤π2
0<cot−1 x≤π
Example 2: Evaluate
(a)
(b)
sin ( tan−1√3)
sec (cos−1−0.5)
The Inverse Trigonometric Functions
−π2
≤sin−1 x≤π2
0≤cos−1 x≤π
−π2
≤csc−1 x≤π2
csc−1 x≠0
0≤sec−1 x≤πsec−1 x≠π
2
−π2
< tan−1 x≤π2
0<cot−1 x≤π
Example 3: Evaluate sin (2 tan−1 2)
The Inverse Trigonometric Functions
−π2
≤sin−1 x≤π2
0≤cos−1 x≤π
−π2
≤csc−1 x≤π2
csc−1 x≠0
0≤sec−1 x≤πsec−1 x≠π
2
−π2
< tan−1 x≤π2
0<cot−1 x≤π
Example 4: Find the exact value of x for
tan−1 x=sin−1(25)
The Inverse Trigonometric Functions
−π2
≤sin−1 x≤π2
0≤cos−1 x≤π
−π2
≤csc−1 x≤π2
csc−1 x≠0
0≤sec−1 x≤πsec−1 x≠π
2
−π2
< tan−1 x≤π2
0<cot−1 x≤π
Example 5: Solve the equation
for x.1− y=cos−1(1−x)
