Download Unit 5 Trigonometric Identities

Precalculus Notes: Unit 5 – Trigonometric Identities
Syllabus Objectives: 3.3 – The student will simplify trigonometric expressions and prove
trigonometric identities (fundamental identities). 3.4 – The student will solve trigonometric
equations with and without technology.
Identity: a statement that is true for all values for which both sides are defined
Example from algebra: 3 x 8 11 3 x 13
sin
Reciprocal Identities
csc
Quotient Identities
tan
Recall: Unit Circle
r 1, x
1
csc
1
sin
sec
sin
cos
cos , y
1
sec
1
cos
cos
cot
cos
sin
cot
or cos
sin
1
cot
1
tan
tan
x
1
x
y
1
sin
y
x, y
Note: sin 2
Pythagorean Theorem: x 2
y2
1
sin
Pythagorean Identity: sin 2
To derive the other Pythagorean Identities, divide the entire equation by sin 2
sin 2
sin 2
cos 2
sin 2
1
sin 2
Pythagorean Identities
1 cot 2
sin 2
csc 2
cos 2
2
sin 2
cos 2
1 1 cot 2
cos 2
cos 2
csc 2
and then by cos 2
1
cos 2
tan 2
cos 2
tan 2
1
1
:
1 sec 2
sec 2
Simplifying Trigonometric Expressions:
Look for identities
Change everything to sine and cosine and reduce
Note that the equations in bold are the trig identities used when simplifying. All of the other steps are
algebra steps.
Ex1: Use basic identities to simplify the expressions.
cos
cot
sin 2
a) cot 1 cos2
sin
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cos 2
1
sin 2
1 cos 2
Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
Precalculus Notes: Unit 5 – Trigonometric Identities
cos
sin
1 cos 2
cot
b) tan
csc
tan
csc
cos
sin
sin 2
1
sin
csc x 1 csc x 1
2
csc x 1
cos 2 x
cos 2 x
cos 2 x
sin 2 x
cos 2 x
2
cot x
cos 2 x
2
.
cos 2 x
csc 2 x 1
cos 2 x
1 cot 2
1
sin 2 x
2
sin x cos x
1
sin
sec
csc x 1 csc x 1
cos 2 x
cos sin
csc
1
cos
Ex2: Simplify the expression
Use algebra:
sin
sin
cos
tan
sin
cos
sin
csc 2
cot 2
csc 2
1
csc 2 x
Solving Trigonometric Equations
Isolate the trigonometric function.
Solve for x using inverse trig functions. Note – There may be more than one solution or no
solution.
Ex3: Solve the equation 4 sin 2 x 4
4sin 2 x 4
sin 2 x
0
sin 2 x
1
Find values of x for which x
0 in the interval 0, 2
1
sin 1 1 and x
sin x
sin
1
1 :
.
1
x
3
2 2
,
Exploration: Consider a right triangle.
Φ
c
a
θ
b
Note that
and
sin
cos
csc
sec
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are complementary. Write the trig functions for each angle. What do you notice?
b
c
c
b
cos
sin
sec
csc
a
c
c
a
tan
cot
cot
tan
b
a
c
b
Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
Precalculus Notes: Unit 5 – Trigonometric Identities
are equal to the cofunctions of θ, when
**The trig functions of
and
are complementary.
Cofunction Identities:
sin 90
cos
or sin
cos 90
sin
or cos
sec 90
csc
or sec
csc 90
sec
or csc
tan 90
cot
or cot
cot 90
tan
or cot
cos
2
sin
2
csc
2
sec
2
tan
2
tan
2
Exploration: Consider an angle, θ, and its opposite, as shown in the coordinate grid. Compare the trig
functions of each angle.
z
θ x
−θ
z
y
sin
y
, sin
z
y
z
cos
csc
z
, csc
y
z
y
sec
x
, cos
z
z
, sec
x
**Cosine and secant are the only EVEN f
f
x
f x
x
z
z
x
tan
y
, tan
x
y
x
cot
x
, cot
y
x
y
trig functions. All the rest are ODD
f x .
x
Odd-Even Identities:
sin
sin
and csc
csc
tan
tan
and cot
cot
cos
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cos
and sec
sec
Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
Precalculus Notes: Unit 5 – Trigonometric Identities
Ex4: Simplify the expression sin
sin
x csc
x
sin x
csc x
sin x
x .
x csc
1
sin x
1
sin
x
sin x
csc
x
csc x
Simplifying Trigonometric Expressions: Simplify using the following strategies. Note that the equations
in bold are the trig identities used when simplifying. All of the other steps are algebra steps.
Ex5: Simplify the expression by factoring. cos3 x cos x sin 2 x
cos3 x cos x sin 2 x
cos x cos 2 x sin 2 x
cos x 1
cos x
Ex6: Simplify the expression by combining fractions.
sin x
1 cos x
cos x
sin x
sin x sin x
cos x 1 cos x
1 cos x sin x
sin 2 x cos 2 x cos x
1 cos x sin x
1 cos x
1 cos x sin x
sin x
1 cos x
cos 2 x
1
sin 2 x
cos 2 x
1
csc x
1
sin x
cos x
sin x
sin 2 x cos x cos 2 x
1 cos x sin x
1
sin x
sin 2 x
csc x
Solving Trigonometric Equations: Solve using the following strategies. Find all solutions for each
equation in the interval 0, 2
.
Ex7: Solve the equation by isolating the trig function. 2cos x 1 0
cos x
1
2
x
5
3 3
These are values of x where the cosine is equal to
,
1
.
2
Ex8: Solve the equation by extracting square roots. 4 sin 2 x 3 0
sin 2 x
3
4
sin x
3
2
x
2 4 5
These are values of x where the sine is equal to
,
,
3 3 3 3
,
Ex9: Solve the equation by factoring. 2 cos 2 x cos x 1
2 cos 2 x
Set equal to zero.
Factor.
2cos x 1 0
1
cos x
2
5
x
,
3 3
Note: It may be easier to use u-substitution with u
quadratic equation that can be factored.
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0
2 cos x 1 cos x 1
Set each factor equal to zero.
Solve each equation.
cos x 1
0
cos x 1 0
cos x
x
1
x
3
, ,
5
3
cos x to help students visualize the equation as a
Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
3
.
2
Precalculus Notes: Unit 5 – Trigonometric Identities
Ex10: Solve the equation by factoring. 2sec x sin x sec x
sec x 2 sin x 1 0
Factor out GCF.
Use zero product property.
sec x
Solve each equation.
1
cos x
x
0
0
2sin x 1 0
1
sin x
2
0
x
5
6 6
,
Note: It is possible for an equation to have no solution.
2
Ex11: Solve by rewriting in a single trig function. 2 sin x
2
Substitute Pyth. Identity.
2 1 cos x
Simplify algebraically.
2 2 cos 2 x
Factor and solve.
3cos x
3cos x
Ex12: Solve using trig substitutions.
Rewrite sin 3 x
Rewrite
sin x
cos x
0
sin 3 x
cos x
cos x
tan x
tan x .
sin 2 x tan x
tan x
sin 2 x tan x
tan x
0
0
x
0, ,
cos
sin 2
1
1 cos 2
0
1
, cos x 1
2
x
0,
5
3 3
,
tan x
sin 2 x sin x
cos x
Use the zero product property and solve.
3
2
2 cos 2 x 3cos x 1
sin x sin 2 x .
Set equal to zero and factor.
x
sin
3
3
2 cos x 1 cos x 1
3cos x
2
tan x
sin 2 x
sin x
cos x
tan x
tan x sin 2 x 1
0,
sin x
1
0
x
2
,
3
2
3
2 2
,
Ex13: Find the approximate solution using the calculator. 4cos x 1
Isolate the trig function.
cos x
1
4
To find x, we need to find the inverse cosine of ¼. x
When solving an equation in the interval 0, 2
cos
1
4
1
x
1.318
, be sure to be in Radian mode.
You Try: Make the suggested trigonometric substitution and then use the Pythagorean Identities to write
the resulting function as a multiple of a basic trig function.
4
x2 , x
2 cos
QOD: Explain the relationship between trig functions and their cofunctions.
Page 5 of 23
Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
Precalculus Notes: Unit 5 – Trigonometric Identities
Syllabus Objective: 3.3 – The student will simplify trigonometric expressions and prove
trigonometric identities.
Trigonometric Identity: an equation involving trigonometric functions that is a true equation for all
values of x
Tips for Proving Trigonometric Identities:
1. Manipulate only one side of the equation. Start with the more complicated side.
2. Look for any identities (use all that you have learned so far).
3. Change everything to sine or cosine.
4. Use algebra (common denominators, factoring, etc) to simplify.
5. Each step should have one change only.
6. The final step should have the same expression on both sides of the equation.
Note: Your goal when proving a trig identity is to make both sides look identical!
For all of the following examples, prove that the identity is true. The trig identities used in the
substitutions are in bold.
Ex1: cos3 x
1 sin 2 x cos x
Start with the right side (more complicated).
cos3 x
1 sin 2 x cos x
sin 2 x
cos 2 x cos x
cos3 x
cos 2 x
1
cos 2 x
1 sin 2 x
cos3 x
Ex2:
1
1 sin x
1
1 sin x
2 sec 2 x
Start with the left side.
Combine fractions.
Simplify.
Trig substitution.
Identity.
Page 6 of 23
1
1 sin x 1
1 sin x 1
1 sin x 1
1
2 sec 2 x
sin x
sin x
sin x
2
1 sin 2 x
2
cos 2 x
2 sec 2 x
2 sec 2 x
sin 2 x
cos 2 x
1
cos x
sec x
1
cos 2 x
1 sin 2 x
Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
Precalculus Notes: Unit 5 – Trigonometric Identities
Ex3:
tan 2 x 1 cos2 x 1
tan 2 x
tan 2 x 1 cos2 x 1
Start with the left side.
Trig substitution.
sec 2 x cos 2 x 1
Trig substitution.
sec 2 x
sin 2 x
Trig substitution.
1
cos 2 x
sin 2 x
Multiply.
sin 2 x
cos 2 x
Identitiy.
tan 2 x
Ex4: sec x tan x
tan 2 x 1
sin 2 x
tan 2 x
sec x
tan x
Change to sine/cosine.
1
cos x
sin x
cos x
Combine fractions.
Multiply num/den by conjugate.
1 sin 2 x
cos x 1 sin x
Trig substitution.
cos 2 x
cos x 1 sin x
Simplify.
cos x
1 sin x
1
1
cos x
sec x
sin x
cos x
tan x
1
cos x
sec x
sin 2 x
cos x
1 sin x
sin x
cos x
tan x
sin 2 x
cos 2 x
cos 2 x
1
1 sin 2 x
cos x
1 sin x
sin 2
Start with left side.
sec 2
sec 2
1
Split the fraction.
sec 2
sec 2
1
sec 2
Simplify.
1
Trig substitution.
1 cos 2
Identity.
sin 2
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cos 2 x 1
1
1 sin x
cos x
1 sin x 1 sin x
cos x 1 sin x
Multiply.
sec 2
sec 2
cos 2 x
sec 2 x
cos x
1 sin x
Start with the left side.
Ex5:
tan 2 x
sin 2
1
sec 2
sin 2
1
sec
cos
sin 2
cos 2
1
sin 2
1 cos 2
Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
Precalculus Notes: Unit 5 – Trigonometric Identities
Challenge: Try to prove the identity above in another way.
You Try: Prove the identity.
cos x sin x
2
cos x sin x
2
2
QOD: List at least 5 strategies you can use when proving trigonometric identities.
Page 8 of 23
Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
Precalculus Notes: Unit 5 – Trigonometric Identities
Syllabus Objective: 3.3 – The student will simplify trigonometric expressions and prove
trigonometric identities (sum and difference identities).
Recall:
36
64
100
So in general,
10
a
36
b
a
64
36
64
6 8
14
b
and
2 3
2
5
2
25
2 3
So in general, a b
2
a2
2
2
2
3
2
13
b2
Sum and Difference Identities
sin u
v
sin u cos v
cos u sin v
cos u
v
cos u cos v
sin u sin v
tan u
v
tan u tan v
1 tan u tan v
Note: Be careful with +/− signs!
Simplifying Expressions with Sum and Differences
1. Rewrite the expression using a sum/difference identity.
2. Simplify the expression and evaluate if necessary.
Ex1: Write the expression as the sine of an angle. Then give the exact value.
sin
sin
4
cos
12
sin
cos
4
2
12
4
cos
sin
sin
cos
12
12
sin
4
4
sin
12
sin u v
12
sin u cos v
cos u sin v
1
2
6
Evaluating Trigonometric Expressions with Non-Special Angles
1. Rewrite the angle as a sum or difference of two special angles.
2. Rewrite the expression using a sum/difference identity.
3. Evaluate the expression.
or
Ex2: Find the exact value of cos195 .
1. 195
2.
150
cos u v
Page 9 of 23
45
cos u cos v
cos195
sin u sin v
cos 150
45
cos 150
45
cos150 cos 45
sin150 sin 45
Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
Precalculus Notes: Unit 5 – Trigonometric Identities
3.
cos150 cos 45
3
2
sin150 sin 45
2
2
1
2
Ex3: Write as one trig function and find an exact value.
1.
tan u v
2.
tan 80
3.
tan 135
tan u tan v
1 tan u tan v
55
2
2
6
4
2
4
6
4
2
tan 80 tan 55
1 tan 80 tan 55
tan 80 tan 55
1 tan 80 tan 55
tan 80
55
tan 135
1
Evaluating Trig Functions Given Other Trig Function(s)
Ex4: Find cos u v given cos u
cos u v
cos u cos v
sin u sin v
15
,
17
u
3
and sin v
2
4
,0
5
v
2
.
We must find cos v and sin u .
Draw the appropriate right triangles in the coordinate plane.
cos u
15
,
17
u
3
:
2
4
,0
5
sin v
v
2
:
15
u
5
4
17
v
Use the Pythagorean Theorem to find the missing sides.
152 U 2
U
17 2
V2
8
cos u cos v
Page 10 of 23
sin u sin v
52
V 3
8
. In Quadrant I, cosine is positive, so cos v
17
In Quadrant III, sine is negative, so sin u
cos u v
42
15
17
3
5
8
17
4
5
45
85
32
85
77
85
Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
3
.
5
Precalculus Notes: Unit 5 – Trigonometric Identities
Proving Identities
Ex5: Verify the identity.
sin
tan
cos cos
tan
Start with the left side.
sin
tan
cos cos
Trig substitution:
Split the fraction:
tan
sin u cos v
cos u sin v
sin cos
cos sin
cos cos
sin
cos
cos cos
cos
cos
Simplify:
sin
cos
sin
cos
Trig substitution:
tan
tan
You Try: Verify the cofunction identity sin
sin
cos
tan
2
tan
cos
QOD: Give an example of a function for which f a b
Then give an example of a function for which f a b
Page 11 of 23
sin u v
using the angle difference identity.
f a
f a
f b for all real numbers a and b.
f b for all real numbers a and b.
Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
Precalculus Notes: Unit 5 – Trigonometric Identities
Syllabus Objective: 3.3 – The student will simplify trigonometric expressions and prove
trigonometric identities (double angle and power-reducing identities).
Ex1: Derive the double angle identities using the sum identities.
1.
sin 2u
sin u
sin u
2.
cos 2u
u
u
cos u
cos u
3.
tan 2u
2 sin u cos u
cos u cos u sin u sin u
cos 2 u sin 2 u
u
u
tan u
sin u cos u sin u cos u
u
tan u tan u
1 tan u tan u
tan u u
2 tan u
1 tan 2 u
Double Angle Identities
sin 2
2 sin cos
cos 2
cos 2
sin 2
2 tan
tan 2
1 tan 2
There are two other ways to write the double angle identity for cosine. Use the Pythagorean
identity.
cos 2
cos 2
1 cos 2
cos 2
1 2 sin 2
1 sin 2
cos 2
2 cos 2
sin 2
cos 2
sin 2
cos 2
Page 12 of 23
1
sin 2
1
Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
Precalculus Notes: Unit 5 – Trigonometric Identities
Evaluating Double-Angle Trigonometric Functions
Ex: Find the exact value of cos2u given cot u
5,
3
2
2 .
u
5
u
1
u will be in Quadrant IV and forms a right triangle as labeled.
Using the Pythagorean Theorem, we have c 2
cos 2 u
Double Angle Identity: cos 2 u
2
3
2
2u
2 2
3
3
2
c
26
5
cos u
1
26
Note: If u is in Quadrant IV,
12
sin 2 u
2
5
cos 2u
52
2
25
26
26
26
1
26
, sin u
24
26
1
26
12
13
2 , then for 2u we have
u
4 , which is in Quadrant IV. So it makes sense that cos2u is positive.
2u
Solving Trigonometric Equations
Ex2: Find the solutions to 4sin x cos x 1 in 0, 2
Rewrite the equation.
2 2sin x cos x 1
Trig substitution.
2 sin 2 x
Isolate trig function.
sin 2 x
Solve for the argument.
2x
1
.
sin2 x
1
2
sin
1
1
2
Because the argument is 2x, we must revisit the domain. 0, 2
0
x
2 . Therefore, 2 0
2x
2 2
2x
Solve for x.
Page 13 of 23
x
6
2sin x cos x
0 2x
, 2x
is the restriction for x. So
4 .
5
, 2x
6
13
, 2x
6
17
6
5
13
17
,
,
12 12 12
12
,
Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
Precalculus Notes: Unit 5 – Trigonometric Identities
Rewriting a Multiple Angle Trig Function to a Single Angle
Ex3: Express sin 3x in terms of sin x .
sin 3 x sin 2 x
Rewrite argument as a sum
Sum identity
x
sin x cos2 x sin2 x cos x
sin x 1 2 sin 2 x
Double angle identities
Pythagorean identity
2 sin x cos x cos x
sin x
2 sin 3 x
2 sin x cos 2 x
sin x
2 sin 3 x
2 sin x 1 sin 2 x
sin x
2 sin 3 x
2 sin x
2 sin 3 x
3sin x 4sin 3 x
Simplify
Verifying a Trig Identity
Ex4: Verify sin 2
2 tan
1 tan 2
Start with left side.
sin 2
Pythagorean identity
2 tan
1 tan 2
2 tan
sec 2
sin
2
cos
1
cos 2
Rewrite in sines/cosines
Simplify
sin
cos 2
cos
1
2sin cos
2
Double angle identity
sin2
cos 2
cos 2
sin 2
1 2 sin 2
Recall: cos 2
2 cos 2
cos 2
Solving for sin 2
.
and cos 2
1
, we can derive the power reducing identities.
cos 2
1 2 sin 2
cos 2
2 cos 2
sin 2
1 cos 2
2
cos 2
1 cos 2
2
tan 2
sin 2
cos 2
Page 14 of 23
1 cos 2
2
1 cos 2
2
1
1 cos 2
1 cos 2
Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
Precalculus Notes: Unit 5 – Trigonometric Identities
Power Reducing Identities
sin 2
cos 2
tan 2
1 cos 2
2
1 cos 2
2
1 cos 2
1 cos 2
Ex5: Express cos5 x in terms of trig functions with no power greater than 1.
cos 5 x
Rewrite as a product
cos 2 x cos 2 x cos x
1 cos 2 x 1 cos 2 x
cos x
2
2
1
1 2 cos 2 x cos 2 2 x cos x
4
Power reducing identity
Multiply
1
1 2 cos 2 x
4
Power reducing identity
1 cos 4 x
cos x
2
You Try:
1. Find the solutions to 2cos x sin 2 x
2. Verify
2 cos 2
sin 2
cot
tan
0 in 0, 2
.
.
QOD: How do you convert from a cosine function to a sine function? Explain.
Page 15 of 23
Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
Precalculus Notes: Unit 5 – Trigonometric Identities
Syllabus Objective: 3.3 – The student will simplify trigonometric expressions and prove
trigonometric identities (half angle identities).
Recall: sin 2
1 cos 2
2
Solving for sin
u
u
, we have sin
2
2
u
u
. We have sin 2
2
2
Let
1 cos u
2
1 cos u
. All of the other half-angle identities can be derived
2
in a similar manner.
Half-Angle Identities
sin
u
2
1 cos u
2
cos
u
2
1 cos u
2
tan
u
2
1 cos u
1 cos u
Note: The
u
2
u
tan
2
tan
Note: There are 2 others for tangent.
will be decided based upon which quadrant
1 cos u
sin u
sin u
1 cos u
u
lies in.
2
Evaluating Trig Functions
Ex1: Find the exact value of cos
Rewrite as a half angle
cos
12
.
1
2 6
cos
12
1 cos
Half angle identity
6
2
3
2
1
Evaluate
2
2
3
4
Choose sign
2
3
2
12
is in Quadrant I, where cosine is positive.
Note: This can also be found using a difference identity. cos
Page 16 of 23
12
cos
3
4
Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
Precalculus Notes: Unit 5 – Trigonometric Identities
Solving a Trig Equation
Ex2: Solve the equation sin x
Half-angle identity
sin
x
in 0, 2
2
.
1 cos x
2
sin x
1 cos x
2
1 cos x
1 cos 2 x
2
sin 2 x
Square both sides
Pythagorean identity
2 2 cos 2 x
Set equal to zero
1 cos x
2
2 cos x cos x 1 0
Factor
2 cos x 1 cos x 1
2 cos x 1
0
0
cos x 1 0
1
cos x
2
2
4
x
,
3
3
Zero product property
cos x
x
1
0
x
0,
2 4
,
3 3
You Try:
1. Find the exact value of tan 22.5 .
2. Solve the equation sin 2
x
2
cos x
0 in 0, 2
.
QOD: Explain why two of the half-angle identities do not have +/− signs.
Page 17 of 23
Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
Precalculus Notes: Unit 5 – Trigonometric Identities
Syllabus Objective: 3.5 – The student will solve application problems involving triangles (Law of
Sines).
Deriving the Law of Sines: Consider the two triangles.
C
C
b
A
b
a h
A
c B
a
h
B
c
h
h
and sin B
.
b
a
b sin A and h a sin B
In the acute triangle, sin A
Solve for h.
h
In the obtuse triangle, sin
sin A
a
sin C
The same type of argument can be used to show that
c
Substitute.
a sin B
b sin A can be rewritten as
B
sin B
h
.
a
sin B
b
sin B
b
sin A
.
a
Law of Sines: The ratio of the sine of an angle to the length of its opposite side is the same for all three
angles of any triangle.
C
a
b
sin A
a
sin B
b
sin C
or
c
a
sin A
b
sin B
c
sin C
A
B
c
Solving a Triangle: finding all of the missing sides and angles
Note: The Law of Sines can be used to solve triangles given AAS and ASA.
Ex1: Solve the triangle
ABC given that
B
15 ,
C
52 , and b
9.
C
a
9
A
sin B
b
c
sin C
c
B
We are given AAS, so we will use the Law of Sines.
sin15
9
sin 52
c
Solve for c: c sin15
Find m A using the triangle sum.
m A 180
sin B
b
Solve for a: a sin15
sin A
a
sin15
9
sin113
a
ABC : m A 113 , m B
Page 18 of 23
15 , m C
52 , a
15 52
32, b
9, c
9 sin 52
9 sin 52
sin15
c
27.4
113
9 sin113
c
9 sin113
sin15
27.4
Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
32
Precalculus Notes: Unit 5 – Trigonometric Identities
Ambiguous Case: When given SSA, there could be 2 triangles, 1 triangle, or no triangles that can be
created with the given information.
Ex1: Solve the triangle
ABC (if possible) when m C
sin C
c
Given SSA, use Law of Sines.
sin A
a
54 , a
sin 54
7
7.
10, c
sin A
10
Solve for A.
10sin 54
7sin A
sin A
10sin 54
7
A sin
10sin 54
7
1
sin
1
1.1557
Solution: There is no possible triangle with the given information because there is no angle with a sine
greater than 1.
Ex2: Solve the triangle
ABC (if possible) when m C
sin C
c
Given SSA, use Law of Sines.
sin B
b
31 , b
sin 31
29
46, c
29 .
sin B
46
Solve for B.
46sin 31
29sin B
46sin 31
29
sin B
B
sin
1
46sin 31
29
54.8
Note that the calculator only gives the acute angle measure for B. There does exist an obtuse angle B
with the same sine. m B 180 54.8 125.2 This is also an appropriate measure of an angle in a
triangle, so there are 2 triangles that can be formed with the given information.
Triangle 1
Triangle 2
m A 180
sin C
c
a sin 31
sin A
a
54.8 31
sin 31
29
29 sin 94.2
Triangle 1: m A 94.2 , m B
94.2
m A 180
sin 94.2
a
sin C
c
a
56.2
54.8 , m C
Triangle 2: m A 23.8 , m B 125.2 , m C
Page 19 of 23
sin A
a
a sin 31
31 , a
31 , a
125.2 31
56.2, b
22.7, b
sin 31
29
29 sin 23.8
46, c
46, c
23.8
sin 23.8
a
a
22.7
29
29
Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
Precalculus Notes: Unit 5 – Trigonometric Identities
Application Problems
1. Draw a picture!
2. Use the Law of Sines to solve for what is asked in the problem.
Ex3: The angle of elevation to a mountain is 3.5 . After driving 13 miles, the angle of elevation is
9 . Approximate the height of the mountain.
z
3.5° θ
13
h
9°
First, find θ:
(not drawn to scale)
180
9
171 . Therefore, the third angle in the small triangle = 5.5°.
Using the law of sines, we know that
13
sin 5.5
z
sin171
Now we can use right triangle trig to find h: sin 3.5
You Try: Solve the triangle
sin 5.5 z
h
21.22
ABC (if possible) when m B
13sin171
h
z
21.22 .
1.3miles
98 , b
10, c
3.
QOD: Explain why SSA is the ambiguous case when solving triangles.
Page 20 of 23
Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
Precalculus Notes: Unit 5 – Trigonometric Identities
Syllabus Objective: 3.5 – The student will solve application problems involving triangles (Law of
Cosines).
Law of Cosines: For any triangle, ABC
C
a
b
c
2
b
2
a2
a
2
a
2
b2
b
2
2ab cos C
c
2
2ac cos B
c2
2bc cos A
Note: In a right triangle, c 2
a2
A
b2
B
c
c2
2ab cos 90
a2
b2
2ab 0
c2
a2
b 2 (Pythagorean
Theorem)
The Law of Cosines can be used to solve triangles when given SAS or SSS.
Ex1: Solve the triangle ABC when m A
Note: The given information is SAS. Use a 2
a2
42 2
152
b2
a2
2 42 15 cos 49
49 , b
c2
15 .
42, & c
2bc cos A .
1162.36
a
34.09
Now that we have a matching pair of a side and angle, we can use the Law of Sines.
a
sin A
42
42sin 49
sin B
sin B
34.09
Now find the two possibilities for m C using the triangle sum:
C
b
sin B
180
34.09
sin 49
49 68.4
62.6 or C
180
68.4 or B
B
49 111.6
49 , m B
111.6 , m C
19.4 , a
Ex2: Solve the triangle ABC when a
Note: The given information is SSS. Use a 2
312
52 2
282
2 52 28 cos A
34.09, b
cos A
c2
111.6
19.4 .
42, c 15
28 .
31, b 52, & c
b2
68.4
19.4
Since c is the shortest side, it must be opposite the smallest angle. So m C
m A
180
2bc cos A . (Or any of them!)
2527
2912
A
29.8
Now that we have a matching pair of a side and angle, we can use the Law of Sines.
a
sin A
b
sin B
31
sin 29.8
52
sin B
sin B
52sin 29.8
31
B
56.5 or B
Now find the two possibilities for m C using the triangle sum:
C 180 29.8 56.5 93.7 or C 180 29.8 123.5
180 56.5
26.7
Since c is the shortest side, it must be opposite the smallest angle. So m C
m A
Page 21 of 23
29.8 , m B
123.5 , m C
26.7 , a
31, b
52, c
26.7 .
28
Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
123.5
Precalculus Notes: Unit 5 – Trigonometric Identities
Application Problems
1. Draw a picture!
2. Use the Law of Cosines to solve for what is asked in the problem.
Ex3: A plane takes off and travels 60 miles, then turns 15° and travels for 80 miles. How far is the
plane from the airport?
15°
165°
60
80
c
(not drawn to scale)
Using the picture, we can find the angle in the triangle to give us SAS.
Use the Law of Cosines: c 2
c
2
60
2
80
2
a2
b2
2 60 80 cos165
2 ab cos C
c
2
19272.89
c 138.8 miles
Area of a Triangle
Recall: A
1
bh
2
h
c
sin
c
h
h
c sin ;
A
1
bc sin
2
h
θ
b
b
Formula for the Area of a Triangle Given SAS:
1
ab sin C
2
A
Ex: Find the area of triangle ABC shown.
B
120°
50
32
C
A
A
1
32 50 sin120
2
1
ac sin B .
2
We will use A
A
16 50
3
2
A
400 3 sq units
Heron’s Area Formula
Semi-Perimeter: s
a
b c
2
Area of a Triangle Given SSS: A
Page 22 of 23
s s
a
s
b
s
c
Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
Precalculus Notes: Unit 5 – Trigonometric Identities
Ex4: Find the area of the triangle with side lengths 5 m, 6 m, and 9 m.
Semiperimeter: s
A
s s a
A
s b
a
b c
2
s
5 6 9
2
s
10
s c
10 10 5 10 6 10 9
A
10 5 4 1
A
200
A
10 2 m 2
You Try: Two ships leave port with a 19° angle between their planned routes. If they are traveling at 23
mph and 31 mph, how far apart are they in 3 hours?
QOD: Can there be an ambiguous case when using the Law of Cosines? Explain why or why not.
Page 23 of 23
Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4