Transcript Right Triangle Trigonometry

Trigonometric Equations
Section 5.5
Objectives
• Solve trigonometric equations.
Solve the equation on the interval
0, 2 
3
sin(t )  
2
This question is asking “What angle(s) on the interval [0, 2π) have a
sine value of  3 ?”
2
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.

3
2
T
h
i
s
e
q
4
t 
3
5
or t 
3
Solve the equation on the interval
0, 2 
1
cos(t )  
2
This question is asking “What angle(s) on the interval [0, 2π) have a
1
cosine value of 
?”
2
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.
2
t 
3
4
or t 
3
Solve the equation on the interval
0, 2 
1
2
sin (t ) 
2
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.
1
sin(t )  
2
1
sin(t )  
2
sin(t )  
2
2
continued on next slide
Solve the equation on the interval
0, 2 
1
2
sin (t ) 
2
What we have now is really two equations to solve.
sin(t ) 
2
2
or
sin(t )  
2
2
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.
For the left equation:
t 

4
or
3
t 
4
For the right equation:
t 
5
4
or t 
7
4
Solve the equation on the interval
0, 2 
2 sin(t ) cos(t )  cos(t )  2 sin(t )  1  0
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.
2 sin(t ) cos(t )  cos(t )  2 sin(t )  1  0
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.
cos(t )2 sin(t )  1  2 sin(t )  1  0
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
0, 2 
2 sin(t ) cos(t )  cos(t )  2 sin(t )  1  0
2 sin(t )  1cos(t )  1  0
Now this is really two equations that need to be solved
2 sin(t )  1  0
or
cos(t )  1  0
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.
2 sin(t )  1
1
sin(t ) 
2
or
cos(t )  1
continued on next slide
Solve the equation on the interval
0, 2 
2 sin(t ) cos(t )  cos(t )  2 sin(t )  1  0
Each of these equations, we should know the answers to.
sin(t ) 
t 

6
1
2
5
or t 
6
or
or
cos(t )  1
t 
Solve the equation on the interval
0, 2 
sin(2x )  cos( x )  0
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.
2 sin( x ) cos( x )  cos( x )  0
Now we can factor a cos(x) out of each piece to get:
cos( x )(2 sin( x )  1)  0
continued on next slide
Solve the equation on the interval
0, 2 
sin(2x )  cos( x )  0
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.
2 sin( x ) cos( x )  cos( x )  0
Now we can factor a cos(x) out of each piece to get:
cos( x )(2 sin( x )  1)  0
continued on next slide
Solve the equation on the interval
0, 2 
sin(2x )  cos( x )  0
This is really two equations to solve
cos( x )  0
or
2 sin( x )  1  0
The equation on the left, we should know the solutions to. The
equation on the right, we can solve with some manipulation.
2 sin( x )  1
sin( x )  
t 

2
3
or t 
2
or
t 
7
6
1
2
or t 
11
6
Solve the equation on the interval
0, 2 
sin(t )
2
1

16
Our first step in the problem will be to take the square root of
both sides of the equation.
1
sin(t )  
16
1
sin(t )  
4
This will give us two equations to solve.
sin(t ) 
1
4
or
sin(t )  
1
4
Neither of these is a value that we know from our standard angles.
continued on next slide
Solve the equation on the interval
0, 2 
1
2
sin(t )

16
In order to find these angles, we will need to use our inverse
trigonometric functions.
1
t  sin 1  
4
or
 1
t  sin 1   
 4
These are not all of the answers to the question. Let’s
start by looking at the left side.
1
t  sin  
4
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 


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 .
continued on next slide
1
Solve the equation on the interval
0, 2 
1
2
sin(t )

16
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
1
t  sin  
4
1
since this angle is in quadrant I and all angles in quadrant I and in
the interval [0, 2π) is its own reference angle.
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:
reference angle    angle
continued on next slide
Solve the equation on the interval
0, 2 
1
2
sin(t )

16
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.
reference angle    angle
1
sin -1      angle
4
1
sin -1      angle
4
1
- sin -1      angle
4
1
 - sin -1    angle
4
continued on next slide
Solve the equation on the interval
0, 2 
1
2
sin(t )

16
1

4
This gives us the following two values for the solution to t  sin 1 
1
t  sin 1  
4
or
1
t    sin 1  
4
Now we will work on the other equation that was on the
right in a previous slide.
continued on next slide
Solve the equation on the interval
0, 2 
1
2
sin(t )

16
In order to find these angles, we will need to use our inverse
trigonometric functions.
1
t  sin 1  
4
or
 1
t  sin 1   
 4
These are not all of the answers to the question. We will
continue by looking at the right side
 1
t  sin   
 4
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 


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.
continued on next slide
1
Solve the equation on the interval
0, 2 
1
2
sin(t )

16
One thing that we can do to find an angle in the interval [0, 2π),
 1
 that is in the
 4
it to find an angle coterminal to t  sin  
1
interval [0, 2π). We do this
or by adding 2π to the angle we have.
 1
t  sin     2
 4
1
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 ?
continued on next slide
Solve the equation on the interval
0, 2 
1
2
sin(t )

16
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.


 1
reference angle  2   sin 1     2 
 4


 1
reference angle  2  sin 1     2
 4
 1
reference angle   sin 1   
 4
continued on next slide
Solve the equation on the interval
0, 2 
1
2
sin(t )

16
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:
reference angle  angle - 
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.
reference angle  angle - 
 1
- sin -1     angle - 
 4
 1
 - sin -1     angle
 4
continued on next slide
Solve the equation on the interval
0, 2 
1
2
sin(t )

16
1
This gives us the following two values for the solution to t  sin 1   
 4
 1
t  sin 1     2
 4
or
 1
t    sin 1   
 4
Thus we have all solutions to the original equation.
1
t  sin  
4
 1
t  sin     2
 4
1
t    sin  
4
 1
t    sin 1   
 4
1
or
1
1
or
Find all solutions to the equation
cos( x )(2 sin( x )  1)  0
Once factored, this is really two equations to solve
cos( x )  0
or
2 sin( x )  1  0
The equation on the left, we should know the solutions to. The
equation on the right, we can solve with some manipulation.
2 sin( x )  1
sin( x )  
t 

2
3
or t 
2
or
t 
7
6
1
2
or t 
11
6
Solve the equation on the interval
0, 2 
tan( x )
2
 0.9 tan( x )  4  0
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.
 0.9  0.92  4(1)(4)
tan( x ) 
2(1)
 0.9  0.81  16
2
 0.9  16.81
tan( x ) 
2
 0. 9  4 . 1
tan( x ) 
2
tan( x ) 
continued on next slide
Solve the equation on the interval
0, 2 
tan( x )
2
 0.9 tan( x )  4  0
This is really two equations to solve.
 0.9  4.1
2
3.2
tan( x ) 
2
tan( x )  1.6
tan( x ) 
 0.9  4.1
2
5
tan( x ) 
2
tan( x )  2.5
tan( x ) 
We are going to go through the same process that we did with the
previous problem to find the answers.
continued on next slide
Solve the equation on the interval
0, 2 
tan( x )
2
 0.9 tan( x )  4  0
In order to find these angles, we will need to use our inverse
trigonometric functions.
t  tan 1 1.6
or
t  tan 1  2.5
These are not all of the answers to the question. Let’s
start by looking at the left side.
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 


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 .
continued on next slide
t  tan 1 1.6
Solve the equation on the interval
0, 2 
tan( x )
2
 0.9 tan( x )  4  0
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
t  tan 1 1.6
since this angle is in quadrant I and all angles in quadrant I and in
the interval [0, 2π) is its own reference angle.
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:
reference angle  angle - 
continued on next slide
Solve the equation on the interval
0, 2 
tan( x )
2
 0.9 tan( x )  4  0
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.
reference angle  angle  
tan -1 1.6  angle  
tan -1 1.6    angle
continued on next slide
Solve the equation on the interval
0, 2 
tan( x )
2
 0.9 tan( x )  4  0
This gives us the following two values for the solution to t  tan 1 1.6
t  tan 1 1.6
or
t  tan 1 1.6  
Now we will work on the other equation that was on the
right in a previous slide.
continued on next slide
Solve the equation on the interval
0, 2 
tan( x )
2
 0.9 tan( x )  4  0
In order to find these angles, we will need to use our inverse
trigonometric functions.
t  tan 1 1.6
or
t  tan 1  2.5
These are not all of the answers to the question. We will
continue by looking at the right side
t  tan 1  2.5 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 


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.
continued on next slide
Solve the equation on the interval
0, 2 
tan( x )
2
 0.9 tan( x )  4  0
One thing that we can do to find an angle in the interval [0, 2π),
it to find an angle coterminal to t  tan 1  2.5 that is in the
interval [0, 2π). We do this by adding 2π to the angle we have.
t  tan 1  2.5  2
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 ?
continued on next slide
Solve the equation on the interval
0, 2 
tan( x )
2
 0.9 tan( x )  4  0
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.
reference angle  2  tan 1  2.5  2 
reference angle  2  tan 1  2.5  2
reference angle   tan 1  2.5
continued on next slide
Solve the equation on the interval
0, 2 
tan( x )
2
 0.9 tan( x )  4  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:
reference angle    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.
reference angle    angle
 tan -1  2.5     angle
 tan -1  2.5     angle
tan -1  2.5    angle
continued on next slide
Solve the equation on the interval
0, 2 
tan( x )
2
 0.9 tan( x )  4  0
This gives us the following two values for the solution to t  tan 1  2.5
t  tan 1  2.5  2
or
t    tan 1  2.5
Thus we have all solutions to the original equation.
t  tan 1 1.6
or
t    tan 1 1.6
t  tan 1  2.5  2
or
t    tan 1  2.5
continued on next slide