Transcript Trigonometric Equations

Trigonometric Equations
M 140 Precalculus
V. J. Motto
1
sin   
2
7 11

or
6
6
Comments
We begin by looking at solutions using the unit
circle. This give you an opportunity to see that
there are multiple solutions.
Some of these examples will require a bit of algebra.
We will also use the calculator to find solutions.
The calculator will be your primary method for
solving trigonometric equations.
A Review of the Unit Circle
1
sin   
2
We want to solve the equation:
(Unit Circle Solution)
Where on the unit circle is the sine value - 1/2?
But if we want ALL
solutions we could go
another loop around
the unit circle and
come up with more
answers. A loop
around the circle is 2
so if we add 2 to our
answers we'll get more
answers. We can add
another 2 and get
more answers.
sin
7 11

or
6
6
1
All solutions to the equation sin   
would be:
2
7
11

 2k or
 2k where k is an integer
6
6
What this means is as k goes from 0, 1, 2, etc. you would
have the answer with another loop around the unit circle.
Find all solutions to 2cos  3  0
First get the cos  by itself.
3
cos   
2
5
7

or
6
6
What angles
on the unit
circle have this
for a cos
value?
so ALL solutions would be these and however many
loops around the circle you want.
5
7

 2k or
 2k
6
6
We want to solve the equation:
(Calculator Solution)
1
sin   
2
Using our calculator we get:
1
1
1
sin     sin ( )      0.52
2
2
This is an angle in the 4th Quadrant. The
corresponding positive coterminal is 2π – 0.52 or 5.76
radians
This diagram tells us that the sine function is
negative in the 3rd Quadrant. So there is a
solution there. It is π + 0.52 or 3.66
radians.
You can find other solutions by adding
integer multiples 2π to 5.76 or 3.66.
We want to solve the equation:
(Unit Circle Solution)
2cos  3  0
3
First get the cos  by itself. cos   
2
5
7

or
6
6
What angles
on the unit
circle have this
for a cos
value?
so ALL solutions would be these and however many
loops around the circle you want.
5
7

 2k or
 2k
6
6
We want to solve the equation:
(Calculator Solution)
2cos  3  0
3
First get the cos  by itself. cos   
2
3
3
1
cos   
 cos (
)      2.62
2
2
This is an angle in the 2nd Quadrant with a reference
angle of π – 2.62 or 0.52 radians.
This diagram tells us that the cosine function
is negative in the 3rd Quadrant. So there is a
solution there. It is π + 0.52 or 3.66 radians.
You can find other solutions by adding integer
multiples 2π to 5.76 or 3.66.
This would mean only
one loop around the
circle.
1
2
Solve tan   on the interval [0, 2 )
3
Get tan  alone
tan   
2
1
3
1
3
tan   

3
3
What angles on the unit circle have this value for tangent?
Since it can be either plus or
minus, there are 4 values. We
don't add any to go around
again because it says on the
interval from 0 to 2.

 5 7 11
6
,
6
,
6
,
6
You can find the solutions by
using a calculator
2
Solve sin 2  
on the interval [0, 2 )
2
Notice that we
have 2 NOT 
Solve for 
by dividing
by 2
2


8

9
2   2 
4
4
3
11
2 
 2 
4
4

We still ask
where on the
unit circle does
the sine have
this value.
3
 or
4
4
Since you divided the
3
angle in half,  is smaller
or
so you need to take
8
another loop around the
9 circle because you only

want answers between 0
8 and 2 but by the time
11

8
you divide by 2 you'll still
be in that interval.
You will
canbe
find
solutions
by using
a calculator!
The solution
all the
4 values
because
they are
all still in [0, 2)
(If you try another loop around you'll find yourself larger than 2).
If the values that these trig functions equal are NOT exact
values on the unit circle you will need to use your calculator.
Solve cos  0.6 on the interval 0    2
You can use the inverse cosine button on your calculator
(make sure mode is radians) but remember that the range is
only the top half of the unit circle and we want the whole unit
circle so you'll need to figure out the other value from the
one given.
1
cos
0.6  0.93 this value is somewhere in
from calculator:
Quad I
What other quadrant would have the
same cosine value (same x value on
the unit circle)? Quadrant IV
This angle is 2 minus angle from calculator.
2  0.93  5.35
In the next few examples we'll explore
various techniques to manipulate
trigonometric equations so we can solve
them.
We'll find solutions on the interval from 0 to
2. Often written as [0, 2).
The first tip is to try using identities to get in terms of the
same trig function.
cos   sin   sin   0
2
2
Use the Pythagorean Identity to replace this with an
2
equivalent expression using sine.
cos   1  sin 
1  sin   sin   sin   0
2
2
2sin   sin   1  0
2
(2sin   1)(sin   1)  0
1
sin    , sin   1
2
2
Combine like terms,
multiply by -1 and put
in descending order
Factor (think of sin  like x and
this is quadratic)
Set each factor = 0 and solve
7 11 

,
,
6
6 2
When we don't have squared trig functions, we can't use the
Pythagorean identities. If you have two terms with different
trig functions you can try squaring both sides.
cos  sin   0 2
2
Square both sides. Must do whole
side together NOT each term (so
left side will need to be FOILed).
cos 2   2 cos  sin   sin 2   0
re-order terms
sin   cos 12cos
2 cossin
sin  00
2
2
Pythagorean Identity---this equals 1
1  sin 2  0
sin 2  1
3
3
2 
so  
2
4
Double angle Identity
  2sin  cos
sin 2
Get sine term
alone
Where is the sine -1?
Remember to do
another loop when
you have 2
7
7
2 
so  
2
4
HELPFUL HINTS FOR SOLVING
TRIGONOMETRIC EQUATIONS
• Try to get equations in terms of one trig function by using identities.
• Be on the look-out for ways to substitute using identities. This
simplifies many equations.
• Try to get trig functions of the same angle. If one term is cos2 and
another is cos for example, use the double angle formula to express
first term in terms of just  instead of 2
• Get one side equals zero and factor out any common trig functions
• See if equation is quadratic in form and will factor. (replace the trig
function with x to see how it factors if that helps)
• If the angle you are solving for is a multiple of , don't forget to add
2 to your answer for each multiple of  since  will still be less than
2 when solved for.
There are some equations that can't be solved by hand
and we must use a some kind of technology.
Use a graphing utility to solve the equation. Express any solutions
rounded to two decimal places.
22x 17 sin x  3
Graph this side as y1
in your calculator
Graph this side as y2
in your calculator
You want to know where they are equal. That would be
where their graphs intersect. You can use the trace
feature or the intersect feature to find this (or these)
points (there could be more than one point of
intersection).
22x 17 sin x  3
This is off a little due
check: 22 .53  17sin .53  3.066  3 to the fact we
approximated. If you
carried it to more
decimal places you'd
have more accuracy.
After seeing the initial
graph, lets change the
window to get a better
view of the intersection
point and then we'll do
a trace.
Rounded to 2 decimal places, the
point of intersection is x = 0.53