Trigonometric Equations Section 5.5. Objectives Solve trigonometric equations.

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 ?”


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.


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.


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


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


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


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



21

)sin( t

2,0

65

or6

tt

Each of these equations, we should know the answers to.

t


1)cos( tor

or


0)cos()2sin( xx

2,0

0)cos()cos()sin(2 xxx


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:


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


41

)sin(

161

)sin(

t

t

2,0

41

)sin( t


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


2,

2

2,0

41

sin 1t


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:


angleanglereference

2,0

41

sin 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 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.


angle41

sin-

angle41

sin-

angle41

sin

angle41

sin

angleanglereference

1-

1-

1-

1-

2,0


161

)sin( 2 t


41

sin 1t


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


2,

2

2,0

41

sin 1t



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


241

sin 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.

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


41

sinanglereference

241

sin2anglereference

241

sin2anglereference

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 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:


-angleanglereference

2,0


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-


1-

1-


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.


04)tan(9.0)tan( 2 xx

2,0


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.



2,0

6.1)tan(22.3

)tan(

21.49.0

)tan(

x

x

x


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.



2,0


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


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



2,0


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:


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.



2,0



angle6.1tan

angle6.1t

angleanglereference

1-

1-

an



2,0


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



2,0


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


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



2,0


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 ?



2,0


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.



2,0


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


angle5.2tan

angle5.2tan

angle5.2tan

angleanglereference

1-

1-

1-



2,0


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

Trigonometric Equations Section 5.5. Objectives Solve trigonometric equations.

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Transcript of Trigonometric Equations Section 5.5. Objectives Solve trigonometric equations.