Trigonometric Equations Solving for the angle (The second of two note days and a work day) (6.2)(2)
Trigonometric Equations
Solving for the angle
(The second of two note days and a work day)
(6.2)(2)
POD
True/ false:
The x-intercepts of y = sin 2x are ±πn/2.
POD
True/ false:
The x-intercepts of y = sin 2x are π/2 ±πn/2.
There are couple of ways to do this.
1. Set sin 2x = 0, and solve.
2. Consider the x-intercepts of the graph of y = sin x. We’ve changed something in the equation—how does that change the graph? What x-scale could help see it?
Pick up from last time
Solve by factoring.
What are the four steps again?
0tantansin4 2 xxx
Pick up from last time
Step one: Isolate trig functions by factoring.
0)1sin2)(1sin2(tan
0)1sin4(tan
0tantansin42
2
xxx
xx
xxx
Pick up from last time
Start step 2: use inverse trig
functions.
0)1sin2)(1sin2(tan
0)1sin4(tan
0tantansin42
2
xx
xx
xxx
0
0tan
x
x
6
2
1sin
1sin2
01sin2
x
x
x
x
62
1sin
2
1sin
1sin2
01sin2
1
x
x
x
x
Pick up from last time
Step two: Find anglesin one rotation.
Note: the inverse trig function for the last one gives us a value outside the interval 0 ≤ θ ≤ 2π. If we use that tool, we have to build on it.
x
x 0tan
6
5,
6
2
1sin
x
x
6
11,
6
7
62
1sin
2
1sin
1
x
x
x
Pick up from last time
Step three: Find the general solution (all angles).
We can combine those last four into this.
Or this.
nnx
6
5,
6
nx nnx 2
6
5,2
6 nnx
26
11,2
6
7
nx
6
nx
nx
Solving for the variable
Solve for u. (Note, this is not 2 csc4 u.) You could replace the 2u with x or θ for now, if it helps.
042csc4 u
Solving for the variable
Step one: Isolate trig functions by factoring.
The second factor does not provide a meaningful solution.
0)22)(csc22(csc
042csc22
4
uu
u
2
2
2
12sin
22sin
1
22csc
22csc2
u
u
u
u
22csc
22csc2
u
u
Solving for the variable
Step two: Find angles in one rotation. Again, we build off of the angle we get for the negative sine value.
2
22sin u
4
3,
42
2
2sin2 1
u
u
4
7,
4
52
2
2sin2 1
u
u
Solving for the variable
Step three: Find all angles.
Combined, what would the statement be?
2
22sin u
nnu
u
24
3,2
42
4
3,
42
nnu
u
24
7,2
4
52
4
7,
4
52
Solving for the variable
Step three: Find all angles.
Combined, what would the statement be?
We can use this to solve for u. We can also use the separate statements to solve for u. We’ll get the same answer.
nu24
2
Solving for the variable
Step four: Solve for u with all statements.
This is a mouthful. What’s the easier way to present it?
nnu
u
nnu
u
8
3,
8
8
3,
8
24
3,2
42
4
3,
42
nnu
u
nnu
u
8
7,
8
58
7,
8
5
24
7,2
4
52
4
7,
4
52
Solving for the variable
An elegant combination:
We could get this using that single statement from before.
nnu
u
24
3,2
42
4
3,
42
nnu
u
24
7,2
4
52
4
7,
4
52
nu
nu
48
242
Approximating solutions
Group and factor this to start.
Work in degrees for a switch.
06sin3tan10tansin5
Approximating solutions
Step one: Group and factor this to start.
0)2)(sin3tan5(
0)2(sin3)2(sintan5
0)6sin3()tan10tansin5(
06sin3tan10tansin5
Approximating solutions
Step two: Find angles in one rotation.
Once again, we have a factor that does not provide a meaningful solution.
0)2)(sin3tan5(
329180149
14918031
315
3tan
5
3tan
3tan5
1
2sin
Approximating solutions
Step three: Find all angles.
0)2)(sin3tan5(
n180149
Using a graph
Graph to find the roots of this equation in one rotation.
What patterns do you see?
What x-scale could help to see the intercepts?
03sin2sinsin xxx
Using a graph
Graph to find the roots of this equation in one rotation.
Now graph it -2π to 2π. What do you see?
2,
2
3,
3
4,,
3
2,
2,0
28.6,71.4,19.4,14.3,09.2,57.1,0
03sin2sinsin
x
x
xxx
Using a graph
What about this one? How does it compare to the one before?
13sin2sinsin xxx
Using a graph
What about this one? x = .171, 1.24.What do you see here?
If we rewrote the equation to set one side to 0, what would we have done to the graph?
13sin2sinsin xxx
Application
From example 9, page 476.
What do the various elements of this equation signify?
12)79(365
2sin3)(
ttD
Application
From example 9, page 466.
How many days have more than 10.5 hours?
What’s a fast way to find out?
12)79(365
2sin3)(
ttD
Application
From example 9, page 466.
How many days have more than 10.5 hours?
Graph it first on calculators and calculate intersections.
12)79(365
2sin3)(
ttD
Application
From example 9, page 466.
How many days have more than 10.5 hours?
Graph it first on calculators and calculate intersections.
x = 48.6 and 291.9
Now, algebraically.
12)79(365
2sin3)(
ttD
Application
How many days have more than 10.5 hours?
The expression is actually the angle.
5.)79(365
2sin
5.1)79(365
2sin3
5.1012)79(365
2sin3
t
t
t
)79(365
2t
Application
The expression is actually the angle.
Substitute θ for this expression, and solve for θ.
sinθ = -.5
θ = -π/6 ≈ -.52 θ = π/6 + π ≈ 3.67
Un-substitute and solve for t.
)79(365
2t
Application
Un-substitute and solve for t.
Find the difference between the days to answer the question (finally). Graphing was easier…
8.48
21.302
36552.79
52.)79(365
2
t
t
t
2.292
2.2132
36567.379
67.3)79(365
2
t
t
t