Download Section 6.1 Inverse Trigonometric Functions

Section 6.1 Inverse Trigonometric Functions
Recall: If we have a function y = f (x ) then the inverse function y = f −1 ( x ) is that
function which has the property: f −1 ( f ( x )) = f ( f −1 ( x )) = x . In some sense in undoes
what the original function had done. One important property of the inverse function is
that it is a function, in other words for every x you put into the function you get out
exactly one y value.
We already learned how to solve for the inverse of a function in precalculus and there are
two ways to do it. The easiest thing to do is to look at the graph and reflect it across the
line y = x. The harder way to solve for the inverse of a function is to switch the x and y
coordinates and solve for y.
We can find inverse functions of the sine and cosine and we will use the graphing
method.
The first thing we need to do is to restrict the domain so that when we reflect our function
across the line y = x we end up with a function. Consider the basic graph of sine (Figure
1):
1.5
y = sin x
1
0.5
-π
-π/2
0
π/2
π
-0.5
-1
-1.5
Figure 1
The sine takes on y values in the range: -1 ≤ y ≤ 1 so when we find the inverse function
the domain must be the same. The area highlighted in red covers the appropriate span
-π/2≤ x ≤ π/2.
Now all we need to do is to reflect this red area across the line y = x to get the inverse
sine function. This function is called the arcsine or y = arcsin x. The graph of the arcsine
is in Figure 2 at the top of the next page.
Note: With the restriction we have put on the arcsine, it is ONLY defined in quadrants I
and IV so all your answers for acrsine must lie between -π/2 and π/2.
y = sin x
y = arcsin x
2
1.5
1
-π/2
π/2
0.5
0
-0.5
-1
-1.5
-2
Figure 2
y = arcsin x has range -π/2 ≤ y ≤ π/2
The cosine is similar but in this case we will restrict the domain to 0 ≤ x ≤ π because this
also gives us all the y values between –1 and 1. This is shown in the highlighted red area
of the cosine function (see Figure 3)
1.5
y = cos x
1
0.5
π
-π
2π
0
-2π
-0.5
-1
-1.5
Figure 3
Again we reflect this red area across the line y = x to get the inverse cosine function. This
function is called the arcsine or y = arccos x. We can see this in the next graph (Figure
4).
Note: With the restriction we have put on the arccosine, it is ONLY defined in quadrants
I and II so all your answers for acrcosine must lie between 0 and π.
3.5
y = cos x
y = arccos x
3
2.5
2
1.5
1
0.5
π
0
-0.5
-1
-1.5
Figure 4
For all of these inverse functions you can read them the following way:
y = arcsin x : y is the angle whose sine is x.
In other words sin x gives you a ratio and arcsin x gives you an angle.
y = arccos x : y is the angle whose cosine is x.
In other words cos x gives you a ratio and arccos x gives you an angle.
y = arctan x : y is the angle whose tangent is x.
In other words tan x gives you a ratio and arctan x gives you an angle.
Note: The arctangent is only defined from -π/2≤ x ≤ π/2.
On your calculator these functions are not displayed as arc functions. Writing arccosine
and arcsine takes up very much space so we most often use a short hand version of the
inverse:
arcsin x = sin −1 x
and
arccos x = cos −1 x
These two notations will be used interchangeable in this class and the second form is
used by your calculator. sin −1 x is often found just above sin x but different models of
calculator have it in different places.
 3
 . Find y.
Ex 1: y = arccos 

2


Step 1: Draw a triangle in the appropriate quadrant and label the sides. Arccosine is
3
defined in quadrant I and II and
is positive so we draw the triangle in quadrant I
2
(see Figure 5).
Step 2: Identify the angle in the triangle. Very often it will be one of the special
triangles. In this case we have a 30 – 60 – 90 triangle so our angle is 30° or π/6.
2
y
3
2
Figure 5
Ex 2:
 2
 . Find y.
y = arcsin 

 2 
Ex 3. y = arcsin 0
Ex 4. y = arctan( −1)
Ex 5. tan(arctan ( 25)) =

 7π  
 =
 2 
Ex 6. cos  cos 


 5 
Ex 7. csc  tan −1  −  =
 12 


 3 
Ex 8. tan  sin −1  −  
 4 

Section 6.3: Solving Trigonometric Equations
Solve the equation:
Ex 1: 2 sin x + 1 = 0
Ex 2: 3 cot 2 x − 1 = 0
Ex 3: 2 sin(2 x) = 1
Ex 4: 4 cos 3 x − 3 cos x = 0
Ex 5: Find all solutions on [0, 2π)
2 sin 2 x + 3 sin x + 1 = 0
Ex 6: Find all solutions on [0, 2π)
cos x + sin x tan x = 2
Extra problems:
Ex 1. 2 cos x + 1 = 0
Ex 2. sin x(sin x + 1) = 0
Ex 3. 2 sin 2 2 x = 1
Find all solutions on the interval [0, 2π)
Ex 4. cos 3 x = cos x
Ex 5. 3 tan 2 x + 4 tan x − 4 = 0
Section 6.4 Using Fundamental Identities
Recall these identities:
sin θ
cos θ
1
sec(θ ) =
cos θ
1
sin θ
cos θ
cot(θ ) =
sin θ
csc(θ ) =
tan(θ ) =
Pythagorean identities:
sin 2 (θ ) + cos 2 (θ ) = 1
1 + tan 2 θ = sec 2 θ
1 + cot 2 θ = csc 2 θ
We can find all of the trigonometric functions of an angle without drawing a triangle if
we are given sufficient information and we know the trigonometric identities. By
sufficient information we mean two trigonometric functions of the angle which are not
reciprocals of each other.
Ex 1: tan x =
3
3
and cos x = −
find the other trigonometric functions.
3
2
Since all of our trigonometric functions can be expressed in terms of sine and cosine it is
easiest to find both of those functions first.
3 sin x sin x
=
=
.
3
cos x
3
−
2
We now have an equation to solve for sine:
tan x =
3 sin x
3
3
1
−
=−
=
⇒ sin x =
3
3  2 
2
3
−
2
From here we can solve for the other three by taking reciprocals:
csc x = −2
sec x = −
cot x =
2
2 3
=−
3
3
3
= 3
3
We can also use the identities to take complicated expressions and make them simpler.
This is known as simplification.
Ex 2: Simplify cot 2 x − csc 2 x
In this example we have many terms squared so it is going to be easiest to try to use one
of the Pythagorean identities. In this case we will use 1 + cot 2 θ = csc 2 θ
cot 2 x − csc 2 x = cot 2 x − (1 + cot 2 x ) = cot 2 x − 1 − cot 2 x = −1
So cot 2 x − csc 2 x = −1
Ex 3: Simplify
sec 2 x − 1
sin 2 x
Once again we have squared terms so we need to use a Pythagorean identity:
tan 2 θ = sec 2 θ − 1
sec 2 x − 1 tan 2 x sin 2 x 1
1
=
=
=
= sec 2 x
2
2
2
2
2
sin x
sin x cos x sin x cos x
sec 2 x − 1
= sec 2 x
So
2
sin x
Ex 4: Simplify
1 − sin 2 x
csc 2 x − 1
Sometimes a problem requires factorization as well:
Ex 5: Factor and simplify tan 4 x + 2 tan 2 x + 1
The trick to simplifying this problem to see that it is a quadratic equation in tan 2 x . To
see this more clearly we will do a ”u- substitution”. In this case we will let u = tan 2 x .
Then we get the following:
(tan x )
2
2
(
)
+ 2 tan 2 x + 1
u 2 + 2u + 1 = (u + 1) 2
but we don’t want a solution in u so we have to substitute for u = tan 2 x to get
(tan 2 x + 1) 2 = (sec 2 x ) = sec 4 x
2
Ex 6: Factor and simplify sin 2 x sec 2 x − sin 2 x
Here we will factor the common factor sin 2 x and then apply the identity
tan 2 θ = sec 2 θ − 1
sin 2 x sec 2 x − sin 2 x = sin 2 x(sec 2 x − 1) = sin 2 x tan 2 x
Ex 7: Simplify
1
1
−
sec x + 1 sec x − 1
Here we need to find a common denominator.
Section 6.4b: Verifying Trigonometric Identities
Verify the identity:
1. cot 2 y (sec 2 y − 1) = 1
2. cos x + sin x tan x = sec x
3.
sec x − 1
= sec x
1 − cos x
4.
sec x + tan x
= (sec x + tan x) 2
sec x − tan x
5. cos x −
cos x
sin x cos x
=
1 − tan x sin x − cos x
Extra Problems:
1. tan 2 θ + 4 = sec 2 θ + 3
csc 2 x
2.
= csc x sec x
cot x
3.
1
= csc x − sin x
sec x tan x
4.
cos x cot x
− 1 = csc x
1 − sin x
Section 6.5: Sum and Difference Formulas
The following identities are provided without proof. You DO NOT need to memorize
them for the test, they will be provided. You only need to know how to use them.
Sum and Difference Formulas:
sin(u + v) = sin u cos v + cos u sin v
sin(u − v) = sin u cos v − cos u sin v
cos(u + v) = cos u cos v − sin u sin v
cos(u − v) = cos u cos v + sin u sin v
tan u + tan v
1 − tan u tan v
tan u − tan v
tan(u − v) =
1 + tan u tan v
tan(u + v) =
Note: These formulas are VERY SPECIFIC.
You CAN NOT distribute ANY
trigonometric function:
sin(u + v ) ≠ sin u + sin v
 7π π 
Ex 1: Find the exact value of sin
−  using the difference formula.
3
 6
sin(u − v) = sin u cos v − cos u sin v
7π
π
In this case u =
and v =
v
3
 7π π 
 7π   π 
 7π   π 
So sin
−  = sin  cos  − cos  sin 
3
 6
 6   3
 6   3
To solve this we will draw triangles on the coordinate axes and find the values on the
right hand side of the expression:
Ex 2: Find the exact value of sin(195°).
To solve this we need to find two angles whose sum or difference is 195° and are easy
angles to solve. For example 195° = 225° – 30° or 195° = 135° + 60°
So we could solve this more than one way. We will use the sum formula.
sin(195°) = sin(135° + 60°) = sin(135°) cos(60°) + cos(135°) sin(60°)
Once again we need to draw triangles to solve the right hand side of the equation.
Ex 3: Write the expression as the sine, cosine, or tangent of an angle:
sin(140°) cos(50°) + cos(140°) sin(50°)
Here we will use the sum formula for sine:
sin(140°) cos(50°) + cos(140°) sin(50°) = sin (140° +50°) = sin(190°)
 π   3π 
 π   3π 
Ex 4: Find the exact value cos  cos  − sin  sin 
 16   16 
 16   16 
The next example is very similar to a problem on the test:
Ex 5: Suppose sin u = 5/13 and u is in quadrant I and cos v = -3/5 and v is in quadrant III.
Find cos (u – v)
We need to draw one reference triangle for each angle and then apply the formula:
cos(u − v) = cos u cos v + sin u sin v
Section 6.6: Double-angle and Half-angle Formulas
The following identities are provided without proof. You DO NOT need to memorize
them for the test, they will be provided. You only need to know how to use them.
Double Angle Formulas:
Power Reducing Formulas:
sin(2u ) = 2 sin u cos u
sin 2 u =
1 − cos 2u
2
= 2 cos 2 u − 1
= 1 − 2 sin 2 u
cos 2 u =
1 + cos 2u
2
2 tan u
1 − tan 2 u
tan 2 (u ) =
cos(2u ) = cos 2 u − sin 2 u
tan(2u ) =
1 − cos 2u
1 + cos 2u
Half Angle Formulas:
u
1 − cos u
=±
2
2
u
1 + cos u
cos = ±
2
2
sin
Note: These formulas are VERY SPECIFIC.
trigonometric function:
sin(2u ) ≠ 2 sin u
Ex 1:
17
1
θ
4
Find: sin(2θ )
8
 1  4 
sin(2θ ) = 2 sin θ cos θ = 2

 =
17
 17  17 
You CAN NOT distribute ANY
Ex 2: Solve on the interval [0, 2π)
cos 2 x + sin x = 0
To solve this we need to convert to all sine or all cosine. We will convert the cos 2x term
to sines for this problem using the following identity:
cos(2u ) = 1 − 2 sin 2 u
So
cos 2 x + sin x = 0
1 − 2 sin 2 x + sin x = 0
(1 − sin x )(1 + 2 sin x ) = 0
We have two equations now:
OR
1 − sin x = 0
sin x = 1
1 + 2 sin x = 0
sin x = −1 / 2
x = π/2
x = 7π/6, 11π/6
Ex 3: Rewrite the expression using a double angle formula:
6 cos 2 x − 3
(
)
6 cos 2 x − 3 = 3 2 cos 2 x − 1
= 3 cos 2 x
2
and π/2 ≤ u ≤ π
3
We first need to find u so we have to draw a triangle in the second quadrant:
Ex 4: Find sin 2u if cos u = −
3
u
5
-2
Then we have to use the identity:
4 5
 5  2 
 −  = −
sin(2u ) = 2 sin u cos u = 2

9
 3  3 
Ex 5: Rewrite sin 4 ( x ) using power reducing formulas. We want an expression with no
exponents larger than 1.
sin 4 ( x ) = (sin 2 x )
2
2
 1 − cos 2 x 
=

2


1 − 2 cos 2 x + cos 2 2 x
=
4
We must reduce one more time for cos 2 2 x
 1 + cos 4 x 
1 − 2 cos 2 x + 

2


=
4
Ex 6: Use a half angle formula to find sin 165
1 − cos 330
 330 
sin 165 = sin
=
=+
2
 2 
1−
2
3
2 =
2− 3
4