trigonometric equations section 5.5. objectives solve trigonometric equations
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Trigonometric Equations
Section 5.5
Objectives
• Solve trigonometric equations.
Solve the equation on the interval
23
)sin( t
2,0
This equation is asking what angle(s) t in the interval [0, 2π) has a sine
value of ?
23
This is a sine value that we should recognize as one of our standard angle on the unit circle. Thus we need do no work, but instead just answer the question from memory.
35
or3
4 tt
This question is asking “What angle(s) on the interval [0, 2π) have a
sine value of ?” 23
This question is asking “What angle(s) on the interval [0, 2π) have a
cosine value of ?”
Solve the equation on the interval
21
2,0
This is a sine value that we should recognize as one of our standard angle on the unit circle. Thus we need do no work, but instead just answer the question from memory.
34
or3
2 tt
21
)cos( t
Although we recognize the ½ as a value we know, since the sine function is squared, we first must take the square root of both sides of the equation.
Solve the equation on the interval
22
)sin(
2
1)sin(
21
)sin(
t
t
t
2,0
21
)(sin2 t
continued on next slide
These are sine values that we should recognize as some of our standard angles on the unit circle. Thus we need do no work, but instead just answer the question from memory.
What we have now is really two equations to solve.
Solve the equation on the interval
22
)sin(or22
)sin( tt
2,0
21
)(sin2 t
43
or4
tt 4
7or
45
tt
For the left equation:
For the right equation:
Solve the equation on the interval 2,0
01)sin(2)cos()cos()sin(2 tttt
For this problem, we must do some algebraic work to get to an equation like the previous ones. Here we need to factor. This equation can be factored by grouping.
01)sin(21)sin(2)cos( ttt
01)sin(2)cos()cos()sin(2 tttt
The grouping can be seen here with the red and blue boxes. The part surrounded by the red box has a cos(t) that can be factored out of both pieces.
Now you should notice that the part in the square brackets and the part in the blue box are the same. This means that we can factor this part out. continued on next
slide
Solve the equation on the interval
21
)sin(
1)sin(2
t
t
2,0
01)cos(or01)sin(2 tt
Now this is really two equations that need to be solved
01)cos(1)sin(2 tt
01)sin(2)cos()cos()sin(2 tttt
If we solve each of these equations so that the trigonometric function is on the left and all the numbers are on the right, it will looks just like our previous problems.
1)cos( tor
continued on next slide
Solve the equation on the interval
21
)sin( t
2,0
65
or6
tt
Each of these equations, we should know the answers to.
t
01)sin(2)cos()cos()sin(2 tttt
1)cos( tor
or
Solve the equation on the interval
0)cos()2sin( xx
2,0
0)cos()cos()sin(2 xxx
continued on next slide
In order to do this problem, we need to angles to be the same. The angle in the sine function is 2x. The angle in the cosine function is x. Since we have an identity for the double angle of a sine function, we can replace sin(2x) with the identity. This will give us the same angles in all of the trigonometric functions.
0)1)sin(2)(cos( xx
Now we can factor a cos(x) out of each piece to get:
Solve the equation on the interval
0)cos()2sin( xx
2,0
0)cos()cos()sin(2 xxx
continued on next slide
In order to do this problem, we need to angles to be the same. The angle in the sine function is 2x. The angle in the cosine function is x. Since we have an identity for the double angle of a sine function, we can replace sin(2x) with the identity. This will give us the same angles in all of the trigonometric functions.
0)1)sin(2)(cos( xx
Now we can factor a cos(x) out of each piece to get:
Solve the equation on the interval
0)cos()2sin( xx
2,0
21
)sin(
1)sin(2
x
x
This is really two equations to solve
01)sin(2or0)cos( xx
The equation on the left, we should know the solutions to. The equation on the right, we can solve with some manipulation.
23
or2
tt
611
or6
7 ttor
Solve the equation on the interval
41
)sin(
161
)sin(
t
t
2,0
41
)sin( t
continued on next slide
Our first step in the problem will be to take the square root of both sides of the equation.
41
)sin( t
161
)sin( 2 t
This will give us two equations to solve.
or
Neither of these is a value that we know from our standard angles.
This angle is between 0 and π/2 in quadrant I. We know this because the range of the inverse sine
function is
Solve the equation on the interval
2,
2
2,0
41
sin 1t
continued on next slide
In order to find these angles, we will need to use our inverse trigonometric functions.
41
sin 1t
161
)sin( 2 t
These are not all of the answers to the question. Let’s start by looking at the left side.
or
This means that this value for t is one of our answer in the interval that we need. We are not, however, getting all of the angles where the sine value is ¼ .
41
sin 1t
The next question is “What other quadrant will have a positive sine value?” The answer to this question is quadrant II. In quadrant II reference angles are found as follows:
Solve the equation on the interval
angleanglereference
2,0
41
sin 1t
continued on next slide
How do we find the other angles? We use reference angles and what we know about quadrants. All of the angles that have a sine value of ¼ will have the same reference angle. What is this reference angle? In this case, the reference angle is
161
)sin( 2 t
since this angle is in quadrant I and all angles in quadrant I and in the interval [0, 2π) is its own reference angle.
We can use this formula to find the angle in quadrant II since we know the reference angle. We will just plug in the reference angle and solve for the angle in quadrant II.
Solve the equation on the interval
angle41
sin-
angle41
sin-
angle41
sin
angle41
sin
angleanglereference
1-
1-
1-
1-
2,0
continued on next slide
161
)sin( 2 t
Solve the equation on the interval 2,0
41
sin 1t
continued on next slide
This gives us the following two values for the solution to
41
sin 1t
161
)sin( 2 t
Now we will work on the other equation that was on the right in a previous slide.
or
41
sin 1t
This angle is between -π/2 and 0 in quadrant IV. We know this because the range of the inverse
sine function is
Solve the equation on the interval
2,
2
2,0
41
sin 1t
continued on next slide
In order to find these angles, we will need to use our inverse trigonometric functions.
41
sin 1t
161
)sin( 2 t
These are not all of the answers to the question. We will continue by looking at the right side
or
This means that this value for t is not one of our answer in the interval that we need. We need to do a bit more work to get to the answers.
41
sin 1t
Solve the equation on the interval 2,0
241
sin 1
t
continued on next slide
One thing that we can do to find an angle in the interval [0, 2π),
it to find an angle coterminal to that is in the
interval [0, 2π). We do this by adding 2π to the angle we have.
41
sin 1t
161
)sin( 2 t
This is one angle that fits our criteria. How do we find all other angles on the interval [0, 2π) that also have a sine value of -¼ ?
or
Solve the equation on the interval 2,0
41
sinanglereference
241
sin2anglereference
241
sin2anglereference
1
1
1
continued on next slide
How do we find the other angles? We use reference angles and what we know about quadrants. All of the angles that have a sine value of -¼ will have the same reference angle. What is this reference angle? In this case, we will find the reference angle for the one angle that we have in quadrant IV.
161
)sin( 2 t
The next question is “What other quadrant will have a negative sine value?” The answer to this question is quadrant III. In quadrant III reference angles are found as follows:
Solve the equation on the interval
-angleanglereference
2,0
continued on next slide
161
)sin( 2 t
We can use this formula to find the angle in quadrant III since we know the reference angle. We will just plug in the reference angle and solve for the angle in quadrant III.
angle41
sin-
-angle41
sin-
-angleanglereference
1-
1-
Solve the equation on the interval 2,0
41
sin 1tThis gives us the following two values for the solution to
41
sin 1t
161
)sin( 2 t
Thus we have all solutions to the original equation.
or241
sin 1
t
41
sin 1t
or
241
sin 1
t
41
sin 1t
or
41
sin 1t
Find all solutions to the equation
0)1)sin(2)(cos( xx
21
)sin(
1)sin(2
x
x
Once factored, this is really two equations to solve
01)sin(2or0)cos( xx
The equation on the left, we should know the solutions to. The equation on the right, we can solve with some manipulation.
23
or2
tt
611
or6
7 ttor
This equation is a quadratic equation in tan(x). Since it is not easily seen to be factorable, we can use the quadratic formula to make a start on finding the solutions to this equation.
Solve the equation on the interval
04)tan(9.0)tan( 2 xx
2,0
continued on next slide
21.49.0
)tan(
281.169.0
)tan(
21681.09.0
)tan(
)1(2)4)(1(49.09.0
)tan(2
x
x
x
x
This is really two equations to solve.
Solve the equation on the interval
04)tan(9.0)tan( 2 xx
2,0
6.1)tan(22.3
)tan(
21.49.0
)tan(
x
x
x
continued on next slide
5.2)tan(25
)tan(
21.49.0
)tan(
x
x
x
We are going to go through the same process that we did with the previous problem to find the answers.
Solve the equation on the interval
04)tan(9.0)tan( 2 xx
2,0
continued on next slide
This angle is between 0 and π/2 in quadrant I. We know this because the range of the inverse
tangent function is
2,
2
6.1tan 1t
In order to find these angles, we will need to use our inverse trigonometric functions.
5.2tan 1 t
These are not all of the answers to the question. Let’s start by looking at the left side.
or
This means that this value for t is one of our answer in the interval that we need. We are not, however, getting all of the angles where the tangent value is 1.6 .
6.1tan 1t
Solve the equation on the interval
04)tan(9.0)tan( 2 xx
2,0
continued on next slide
The next question is “What other quadrant will have a positive tangent value?” The answer to this question is quadrant III. In quadrant III reference angles are found as follows:
-angleanglereference
6.1tan 1t
How do we find the other angles? We use reference angles and what we know about quadrants. All of the angles that have a tangent value of 1.6 will have the same reference angle. What is this reference angle? In this case, the reference angle is
since this angle is in quadrant I and all angles in quadrant I and in the interval [0, 2π) is its own reference angle.
Solve the equation on the interval
04)tan(9.0)tan( 2 xx
2,0
continued on next slide
We can use this formula to find the angle in quadrant II since we know the reference angle. We will just plug in the reference angle and solve for the angle in quadrant II.
angle6.1tan
angle6.1t
angleanglereference
1-
1-
an
Solve the equation on the interval
04)tan(9.0)tan( 2 xx
2,0
continued on next slide
6.1tan 1tThis gives us the following two values for the solution to
6.1tan 1t
Now we will work on the other equation that was on the right in a previous slide.
or 6.1tan 1t
Solve the equation on the interval
04)tan(9.0)tan( 2 xx
2,0
continued on next slide
This angle is between -π/2 and 0 in quadrant IV. We know this because the range of the inverse
tangent function is
2,
2
6.1tan 1t
In order to find these angles, we will need to use our inverse trigonometric functions.
5.2tan 1 t
These are not all of the answers to the question. We will continue by looking at the right side
or
This means that this value for t is not one of our answer in the interval that we need. We need to do a bit more work to get to the answers.
5.2tan 1 t
Solve the equation on the interval
04)tan(9.0)tan( 2 xx
2,0
continued on next slide
25.2tan 1 t
One thing that we can do to find an angle in the interval [0, 2π),
it to find an angle coterminal to that is in the
interval [0, 2π). We do this by adding 2π to the angle we have.
5.2tan 1 t
This is one angle that fits our criteria. How do we find all other angles on the interval [0, 2π) that also have a tangent value of
-2.5 ?
Solve the equation on the interval
04)tan(9.0)tan( 2 xx
2,0
continued on next slide
5.2tananglereference
25.2tan2anglereference
25.2tan2anglereference
1
1
1
How do we find the other angles? We use reference angles and what we know about quadrants. All of the angles that have a tangent value of -2.5 will have the same reference angle. What is this reference angle? In this case, we will find the reference angle for the one angle that we have in quadrant IV.
Solve the equation on the interval
04)tan(9.0)tan( 2 xx
2,0
continued on next slide
The next question is “What other quadrant will have a negative tangent value?” The answer to this question is quadrant II. In quadrant II reference angles are found as follows: angleanglereference
We can use this formula to find the angle in quadrant II since we know the reference angle. We will just plug in the reference angle and solve for the angle in quadrant II.
angle5.2tan
angle5.2tan
angle5.2tan
angleanglereference
1-
1-
1-
Solve the equation on the interval
04)tan(9.0)tan( 2 xx
2,0
continued on next slide
5.2tan 1 tThis gives us the following two values for the solution to
5.2tan 1 t
Thus we have all solutions to the original equation.
or 25.2tan 1 t
5.2tan 1 t
or
25.2tan 1 t
6.1tan 1t
or
6.1tan 1t