Trigonometry
Notes on the Inverse Trigonometric Functions.
Inverse Functions: First remember some definitions and properties from College Algebra.
Horizontal Line Test
A function, f (x) , passes the horizontal line test,
if and only if,
all horizontal line intersect the graph of y = f (x) at most once.
One-to-one Functions
A function is one-to-one, abbreviated 1-1,
if and only if,
its graph passes the horizontal line tests.
Invertible Functions
A function, f (x) , is invertible,
if and only if,
it is a one-to-one function.
Composition of Functions
( f g ) (x) = f ( g(x) ) .
Inverse Function
A function, f −1 (x) , is the inverse of f (x) ,
if and only if,
−1
both ( f f ) (x) = x and ( f −1 f ) (x) = x .
The Graph of an Inverse Function
The graph of y = f (x) is the graph of y = f (x) reflected about the line y = x .
−1
Example #1: For f (x) = 4x 2 − 3 , a) is it invertible. If it is invertible, b)find its inverse function. Otherwise, b) make domain
restrictions of the function to break it up into multiple functions that are invertible. Then, c) find the inverses of
these functions.
►
a) Since the graph of f (x) = 4x 2 − 3 is a parabola, it is not 1-1, and thus not invertible.
b) Because the vertex of the parabola is t x = 0 , we would want to break the domain up there, giving us
f1 (x) = 4x 2 − 3, x ≥ 0 , and f 2 (x) = 4x 2 − 3, x ≤ 0
c) To find the expression for the inverses, set y = f (x) , interchange x and y, and solve for y.
y1 = 4x 2 − 3
x≥0
y1 ↔ x
y 2 = 4x 2 − 3
x≤0
y2 ↔ x
x = 4y − 3
2
1
y≥0
x + 3 = 4y
2
1
x+3
= y12
4
± x+3
= y1
2
x +3
= y1 = f1−1 (x)
2
x = 4y 2 2 − 3
x + 3 = 4y 2
y≤0
2
x +3
= y22
4
± x+3
= y2
2
x +3
= y 2 = f 2−1 (x)
2
□
SCC:Rickman
Notes on the Inverse Trigonometric Functions.
Page #1 of 7
The Basic Trigonometric Inverse Functions: Since all of the trigonometric functions are periodic, none of them are invertible without making restrictions on their domains. The
restrictions are done trying to get 1) all of the range of the function, 2) a continuous section of the graph, and 3) as close to 0 as
possible. We can’t always satisfy condition 2, but we do when we can. This lead to the following restricted basic trigonometric
functions below. We’ll talk about the other 3 trigonometric functions later.
The Restricted Basic Trigonometric Functions
Restricted Cosine
Restricted Sine
Restricted Tangent
y = cos(x) on [ 0, π]
y = sin(x) on ⎡⎣ - π2 , π2 ⎤⎦
y = tan(x) on ( - π2 , π2 )
Vertical Asymptotes:
x = - π2 , π2
Thus, there inverses become the following.
The Basic Trigonometric Inverse Functions
Inverse Cosine
Inverse Sine
Inverse Tangent
−1
−1
y = arccos(x) = cos (x)
y = arcsin(x) = sin (x)
y = arctan(x) = tan −1 (x)
Domain = [ -1,1]
Domain = [ -1,1]
Domain = ( - ∞, ∞ )
Range = [ 0, π]
Range = ⎡⎣ - , ⎤⎦
Range = ( - π2 , π2 )
π
2
π
2
Horizontal Asymptotes:
y = - π2 , π2
So essentially, while with the trigonometric functions, you give it an angle and it returns the ratio of 2 sides of a right triangle, for the
inverse trigonometric functions you give them the ratio and it returns the angle.
Also, note that by default the inverse trigonometric functions work only in radians.
Plus, notice the 2 notations for the inverses, i.e. y = arccos(x) = cos −1 (x) . The ARC notation, i.e. y = arccos(x) , comes from the fact
that an angle in radians is the length of arc on the unit circle. Thus, when you evaluate an inverse trigonometric function, you are not
only finding an angle, but also the length of arc. While the inverse notation, i.e. y = cos −1 (x) , is the general notation used for inverse
functions.
Finally, remember that in the inverse notation, the -1 is not an exponent. Therefore, cos −1 (x) ≠ sec(x) . They are different functions.
SCC:Rickman
Notes on the Inverse Trigonometric Functions.
Page #2 of 7
Example #2: Evaluate exactly without a calculator. a) cos −1 ( 12 ) , b) sin −1
( ).
f) tan −1 -
( ) , c) arctan ( 0 ) , d) arccos ( - ) , e) arcsin ( - ) , and
2
2
2
2
3
2
3
3
►
To evaluate the inverse trigonometric functions for basic results, we need to think about the unit circle and the restrictions for that
inverse.
a)
(
1
2
,
3
2
)
b)
(
2
2
,
2
2
)
c)
(1, 0 )
cos −1 ( 12 ) =
π
3
arctan ( 0 ) = 0
( ) = π4
2
2
f)
e)
d)
(-
sin −1
2
2
,
2
2
)
(
( ) = 34π
arccos -
2
2
1
2
,-
( ) = - π3
arcsin -
3
2
3
2
(
)
1
2
,-
3
2
)
( ) = tan ( - ) = tan ( ) = - π6
tan −1 -
3
3
−1
1
3
−1
- 12
3
2
□
Example #3: Evaluate to the 5th decimal place with a calculator. a) arccos ( -0.43) , b) sin −1 ( 3) , and c) tan −1 ( -6 ) .
►
With the exception of part b), you just type in the inverse function making sure your calculator is in radians.
a) arccos ( -0.43) = cos −1 ( -0.43) ≈ 2.01529 .
b) sin −1 ( 3) is undefined since 3 is outside the domain of inverse sine. So you’re calculator should have either gave you an error or an
imaginary number.
c) tan −1 ( - 6 ) ≈ -1.40565 .
□
SCC:Rickman
Notes on the Inverse Trigonometric Functions.
Page #3 of 7
The Inverses of Secant, Cosecant and Cotangent: We find these inverses much the same way as for the basic ones. The main thing to note is that the definition of inverse cotangent can
be different in different textbooks. So we’ll use the one I’m most familiar with.
The Restricted Reciprocal Trigonometric Functions
Restricted Secant
Restricted Cosecant
Restricted Cotangent
π
π
π
π
y = sec(x) on ⎣⎡ 0, 2 ) ∪ ( 2 , π ⎦⎤
y = csc(x) on ⎣⎡- 2 , 0 ) ∪ ( 0, 2 ⎦⎤
y = cot(x) on ( 0, π )
Vertical Asymptote:
x = π2
Vertical Asymptote:
x=0
Vertical Asymptotes:
x = 0, π
Thus, their inverses become the following.
The Reciprocal Trigonometric Inverse Functions
Inverse Secant
Inverse Cosecant
Inverse Cotangent
y = arcsec(x) = sec −1 (x)
y = arccsc(x) = csc −1 (x)
y = arccot(x) = cot −1 (x)
Domain = ( - ∞,-1] ∪ [1, ∞ )
Range = ⎡⎣ 0,
π
2
)∪(
π
2
, π ⎤⎦
Domain = ( - ∞,-1] ∪ [1, ∞ )
Domain = ( - ∞, ∞ )
Range = ⎡⎣- , 0 ) ∪ ( 0, ⎤⎦
Range = ( 0, π )
π
2
Horizontal Asymptote:
y = π2
π
2
Horizontal Asymptotes:
y = 0, π
Horizontal Asymptote:
y=0
Note that the calculators don’t have inverse secant, cosecant, or cotangent built in to them. Thus, we need expressions for evaluating
them.
For secant and cosecant, the inverse and reciprocal properties will give us what we what.
For
y = sec −1 ( x )
sec ( y ) = x
1
cos ( y )
=x
cos ( y ) =
1
x
For
y = csc −1 ( x )
csc ( y ) = x
1
sin ( y )
=x
sin ( y ) =
1
x
y = cos −1 ( 1x )
y = sin −1 ( 1x )
sec −1 ( x ) = cos −1 ( 1x )
csc −1 ( x ) = sin −1 ( 1x )
Unfortunately, this won’t work for cotangent. The values for a negative x wouldn’t be what we want to get the graph above for
cotangent.
SCC:Rickman
Notes on the Inverse Trigonometric Functions.
Page #4 of 7
For inverse cotangent we use the cofunction relation.
For y = cot −1 ( x )
cot ( y ) = x
tan ( π2 − y ) = x
π
2
− y = tan −1 ( x )
π
2
− tan −1 ( x ) = y
π
2
− tan −1 ( x ) = cot −1 ( x )
(
)
( )
Example #4: Evaluate exactly without a calculator. a) sec −1 - 2 , b) arccsc ( -2 ) , and c) arccot - 3 .
►
a)
sec
−1
( - 2 ) = cos ( )
= cos ( - )
−1
−1
=
1
- 2
b)
arccsc ( -2 ) = arcsin ( -21 )
c)
π
=6
2
2
3π
4
( )
( )
arccot - 3 = π2 − tan −1 - 3
= π2 − tan −1
( )
- 3 2
12
= π2 − ( - π3 )
=
3 π+ 2 π
6
=
5π
6
□
Example #5: Evaluate to the 5th decimal place with a calculator. a) sec −1 ( 3) , b) arccsc ( -5 ) , and c) arccot ( - 12 ) .
►
a)
sec −1 ( 3) = cos −1 ( 13 )
b)
arccsc ( -5 ) = arcsin ( -51 )
c)
arccot ( - 12 ) = π2 − tan −1 ( - 12 )
≈ - 0.20136
≈ 1.23096
≈ 2.03444
□
Solving Equations Involving Inverse Trigonometric Functions: Example #6: Find all real solutions for 4sec2 ( θ ) − 4sec ( θ ) + 3 = 0 for θ in [ 0, 2π ) . Round to the thousandths.
►
First since we would rather deal with the basic 3 trigonometric functions, write the equation in terms of cosine, and solve for cosine.
4sec2 ( θ ) − 4sec ( θ ) + 3 = 0
LCD = cos 2 ( θ )
4 cos12 ( θ) − 4 cos1( θ) + 3 = 0
4 − 4 cos ( θ ) + 3cos 2 ( θ ) = 0
3cos 2 ( θ ) − 4 cos ( θ ) + 4 = 0
⎡⎣3cos ( θ ) + 2 ⎤⎦ ⎡⎣cos ( θ ) − 2 ⎤⎦ = 0
cos ( θ ) = - 32 , 2
cos ( θ ) = nd
But, 2is outside range of cosine.
2
3
rd
Since, cosine is negative in both the 2 and 3 quadrants, I will have 2 solutions.
3rd Quadrant
2nd Quadrant
nd
For the 3rd quadrant angle, I subtract the
Arccosine will give the 2 quadrant
angle directly.
arccosine from 2π.
2
cos ( θ ) = - 3
cos ( θ ) = - 32
θ = cos −1 ( - 32 ) ≈ 2.301
θ = 2π − cos −1 ( - 32 ) ≈ 3.983
Thus, θ ≈ 2.301, 3.983 . □
SCC:Rickman
Notes on the Inverse Trigonometric Functions.
Page #5 of 7
Example #7: Find all real solutions for 5 tan 2 ( θ ) + 14 tan ( θ ) − 3 = 0 for θ in [ 0, 2π ) . Round to the thousandths.
►
5 tan 2 ( θ ) + 14 tan ( θ ) − 3 = 0
tan ( θ ) =
⎡⎣5 tan ( θ ) − 1⎤⎦ ⎡⎣ tan ( θ ) + 3⎤⎦ = 0
tan ( θ ) = 15 , -3
1
5
1st Quadrant
tan ( θ ) = 15
3rd Quadrant
tan ( θ ) = 15
θ = tan −1 ( 15 )
θ = π + tan −1 ( 15 )
≈ 0.197
≈ 3.339
tan ( θ ) = -3
For this one we have to remember that tan −1 ( -3) will be in the 4th quadrant, but it will be from - π2 to 0 which is outside our range for
θ. Thus, both angles have to be adjusted.
2nd Quadrant
tan ( θ ) = -3
4th Quadrant
tan ( θ ) = -3
θ = π + tan −1 ( -3)
θ = 2π + tan −1 ( -3)
≈ 1.893
≈ 5.034
Thus, θ ≈ 0.197, 3.339, 1.893, 5.034 . □
Applications: Example #8: For an object that is thrown at an angle off the horizontal of θ0 with a speed of V0, its horizontal velocity, Vx, and its
vertical velocity, Vy, are given by:
Vy = -g t + V0 sin ( θ0 )
Vx = V0 cos ( θ0 )
where t is time, and g is the acceleration due to gravity which is g = 32 sft2 = 9.8 sm2 .
V
The total speed, V, and its angle of trajectory from the horizontal, θ, are found from the right triangle
shown to the right.
Find the formulas for a) V, and b) θ. Also, c) find both V and θ, rounded to the hundreths, after 5sec, if
the object is launched at an angle of 50° above the horizontal with an initial speed of 120 fts . d) Is the
θ
object rising or falling? Assume it doesn’t hit the ground during those 5sec.
►
a)
V 2 = Vx2 + Vy2
tan ( θ ) =
Vy
Vx
θ = tan −1
2
2
θ = tan −1
V = g t − 2g tV0 sin ( θ0 ) + V sin ( θ0 ) + V cos ( θ0 )
2
2
2
0
2
2
0
Vx
b)
V = Vx2 + Vy2
V = ⎡⎣ -g t + V0 sin ( θ0 ) ⎤⎦ + ⎡⎣ V0 cos ( θ0 )⎤⎦
Vy
( )
(
Vy
Vx
-g t + V0 sin ( θ0 )
V0 cos ( θ0 )
)
2
= g 2 t 2 − 2g tV0 sin ( θ0 ) + V02
c)
V = g t − 2g tV0 sin ( θ0 ) + V sin ( θ0 ) + V cos ( θ0 )
2
=
2
2
0
( 32 ) ( 5 )
2
2
2
2
0
− 2 ( 32 )( 5 )120sin ( 50° ) + 120
2
2
≈ 102.88 fts
θ = tan
−1
(
-( 32 )( 5) + (120 ) sin ( 50° )
(120 ) cos ( 50° )
)
d)
It’s falling since θ is negative.
≈ - 41.43°
□
SCC:Rickman
Notes on the Inverse Trigonometric Functions.
Page #6 of 7
SCC:Rickman
Notes on the Inverse Trigonometric Functions.
Page #7 of 7