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Precalculus Notes: Unit 5 – Trigonometric Identities
Page 1 of 23 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
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 8 11 3 13x x
Reciprocal Identities
1 1 1sin cos tan
csc sec cot1 1 1
csc sec cotsin cos tan
Quotient Identities sin cos
tan cotcos sin
Recall: Unit Circle 1, cos , sinr x y
Pythagorean Theorem: 2 2 1x y Pythagorean Identity: 2 2sin cos 1
To derive the other Pythagorean Identities, divide the entire equation by 2sin and then by 2cos :
2 22 2
2 2 2
sin cos 11 cot csc
sin sin sin
2 22 2
2 2 2
sin cos 1tan 1 sec
cos cos cos
Pythagorean Identities 2 2 2 2 2 2sin cos 1 1 cot csc tan 1 sec
Simplifying Trigonometric Expressions:
Look for identities
Change everything to sine and cosine and reduce
Ex: Use basic identities to simplify the expressions.
a) 2cot 1 cos cos
cotsin
2 2 2 2sin cos 1 sin 1 cos
2 2cos coscot 1 cos sin
sin sin
sin sin cos sin
,x y
Note: 22sin sin
Precalculus Notes: Unit 5 – Trigonometric Identities
Page 2 of 23 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
b) tan csc sin
tancos
1
cscsin
sintan csc
1
cos sin 1
seccos
Ex: Simplify the expression
2
csc 1 csc 1
cos
x x
x
.
Use algebra: 2
2 2
csc 1 csc 1 csc 1
cos cos
x x x
x x
2 2 2 21 cot csc cot csc 1
2
2 2 22
2 2 2
coscsc 1 cot cossin
cos cos cos
xx x xx
x x x
2 2sin cosx x
22
1csc
sinx
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.
Ex: Solve the equation 24sin 4 0x in the interval 0,2 .
2 2 24sin 4 0 sin 1 sin 1 sin 1x x x x
Find values of x for which 1 1sin 1 and sin 1x x : 3
,2 2
x
Exploration: Consider a right triangle.
Note that and are complementary. Write the trig functions for each angle. What do you notice?
sin cosb
c cos sin
a
c tan cot
b
a
csc secc
b sec csc
c
a cot tan
c
b
**The trig functions of are equal to the cofunctions of θ, when and are complementary.
a
b
c
θ
Φ
Precalculus Notes: Unit 5 – Trigonometric Identities
Page 3 of 23 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
Cofunction Identities:
Exploration: Consider an angle, θ, and its opposite, as shown in the coordinate grid. Compare the trig functions of each angle.
sin , siny y
z z cos , cos
x x
z z tan , tan
y y
x x
csc , cscz z
y y sec , sec
z z
x x cot , cot
x x
y y
**Cosine and secant are the only EVEN f x f x trig functions. All the rest are ODD
f x f x .
Odd-Even Identities:
sin sin and csc csc
tan tan and cot cot
cos cos and sec sec
sin 90 cos or sin cos2
cos 90 sin or cos sin2
sec 90 csc or sec csc2
csc 90 sec or csc sec2
tan 90 cot or cot tan2
cot 90 tan or cot tan2
x
z
z
y
θ −θ
Precalculus Notes: Unit 5 – Trigonometric Identities
Page 4 of 23 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
Ex: Simplify the expression sin cscx x .
1sin csc sin csc sin 1
sinx x x x x
x
sin sinx x csc cscx 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.
Ex: Simplify the expression by factoring. 2 2cos cos sinx x x
3 2 2 2cos cos sin cos cos sin cos 1 cosx x x x x x x x 2 2sin cos 1x x
Ex: Simplify the expression by combining fractions. sin cos
1 cos sin
x x
x x
2 2
2 2
sin sin cos 1 cossin cos sin cos cos
1 cos sin 1 cos sin 1 cos sin
sin cos cos 1 cos
1 cos sin
x x x xx x x x x
x x x x x x
x x x x
x x
1 cos x1
cscsinsin
xxx
2 2sin cos 1
1csc
sin
x x
xx
Solving Trigonometric Equations: Solve using the following strategies. Find all solutions for each
equation in the interval 0,2 .
Ex: Solve the equation by isolating the trig function. 2cos 1 0x
1 5cos ,
2 3 3x x
These are values of x where the cosine is equal to
1
2.
Ex: Solve the equation by extracting square roots. 24sin 3 0x
2 3 3 2 4 5sin sin , , ,
4 2 3 3 3 3x x x
These are values of x where the sine is equal to
3
2 .
Ex: Solve the equation by factoring. 22cos cos 1x x
Set equal to zero. 22cos cos 1 0x x
Factor. 2cos 1 cos 1 0x x
Set each factor equal to zero. 2cos 1 0x cos 1 0x
Solve each equation.
1cos
25
,3 3
x
x
cos 1x
x
5, ,
3 3x
Note: It may be easier to use u-substitution with cosu x to help students visualize the equation as a quadratic equation that can be factored.
Precalculus Notes: Unit 5 – Trigonometric Identities
Page 5 of 23 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
Ex: Solve the equation by factoring. 2sec sin sec 0x x x
Factor out GCF. sec 2sin 1 0x x
Use zero product property. sec 0x 2sin 1 0x
Solve each equation. 1
0cos xx
1sin
2
50, ,
3 3
x
x
Note: It is possible for an equation to have no solution.
Ex: Solve by rewriting in a single trig function. 22sin 3cos 3x x
Substitute Pyth. Identity. 22 1 cos 3cos 3x x 2 2 2 2sin cos 1 sin 1 cos
Simplify algebraically. 2 22 2cos 3cos 3 2cos 3cos 1 0x x x x
Factor and solve. 12cos 1 cos 1 0 cos , cos 1
2x x x x
50, ,
3x
Ex: Solve using trig substitutions. 3sin
tancos
xx
x
Rewrite 3 2sin sin sinx x x . 2
2sin sin sintan sin tan
cos cos
x x xx x x
x x
Rewrite sin
tancos
xx
x . 2sin tan tanx x x
Set equal to zero and factor. 2 2sin tan tan 0 tan sin 1 0x x x x x
Use the zero product property and solve. tan 0 0,x x 3
sin 1 ,2 2
x x
3
0, , ,2 2
x
Ex: Find the approximate solution using the calculator. 4cos 1x
Isolate the trig function. 1
cos4
x
To find x, we need to find the inverse cosine of ¼. 1 1cos
4x
1.318x
When solving an equation in the interval 0,2 , 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. 24 , 2cosx x
QOD: Explain the relationship between trig functions and their cofunctions.
Precalculus Notes: Unit 5 – Trigonometric Identities
Page 6 of 23 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
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: 3 2cos 1 sin cosx x x
Start with the right side (more complicated).
3 2
2
3 3
cos 1 sin cos
cos cos
cos cos
x x x
x x
x x
2 2 2 2sin cos 1 cos 1 sinx x x x
Ex2: 21 12sec
1 sin 1 sinx
x x
Start with the left side.
Combine fractions.
21 12sec
1 sin 1 sin1 sin 1 sin
1 sin 1 sin
xx xx x
x x
Simplify. 2
2
1 sin x
Trig substitution. 2
2
cos x 2 2 2 2sin cos 1 cos 1 sinx x x x
Identity. 2 22sec 2secx x 1
seccos
xx
Ex3: 2 2 2tan 1 cos 1 tanx x x
Start with the left side. 2 2 2tan 1 cos 1 tanx x x
Precalculus Notes: Unit 5 – Trigonometric Identities
Page 7 of 23 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
Trig substitution. 2 2sec cos 1x x 2 2tan 1 secx x
Trig substitution. 2 2sec sinx x 2 2 2 2sin cos 1 cos 1 sinx x x x
Trig substitution. 22
1sin
cosx
x
1sec
cosx
x
Multiply. 2
2
sin
cos
x
x
Identitiy. 2 2tan tanx x sin
tancos
xx
x
Ex4: cos
sec tan1 sin
xx x
x
Start with the left side. cos
sec tan1 sin
xx x
x
Change to sine/cosine. 1 sin
cos cos
x
x x
1sec
cosx
x
sintan
cos
xx
x
Combine fractions. 1 sin
cos
x
x
Multiply num/den by conjugate. 1 sin 1 sin
cos 1 sin
x x
x x
Multiply.
21 sin
cos 1 sin
x
x x
Trig substitution.
2cos
cos 1 sin
x
x x
2 2 2 2sin cos 1 cos 1 sinx x x x
Simplify. cos cos
1 sin 1 sin
x x
x x
Ex5: 2
22
sec 1sin
sec
Start with left side. 2
22
sec 1sin
sec
Split the fraction. 2
2 2
sec 1
sec sec
Simplify. 2
11
sec
Trig substitution. 21 cos 1
cossec
Identity. 2 2sin sin 2 2 2 2sin cos 1 sin 1 cos
Challenge: Try to prove the identity above in another way.
Precalculus Notes: Unit 5 – Trigonometric Identities
Page 8 of 23 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
You Try: Prove the identity. 2 2cos sin cos sin 2x x x x
QOD: List at least 5 strategies you can use when proving trigonometric identities.
Precalculus Notes: Unit 5 – Trigonometric Identities
Page 9 of 23 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
Syllabus Objective: 3.3 – The student will simplify trigonometric expressions and prove trigonometric identities (sum and difference identities).
Recall: 36 64 100 10 36 64 36 64 6 8 14
So in general, a b a b and
2 22 3 5 25 2 2 2
2 3 2 3 13
So in general, a b a b Sum and Difference Identities
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.
Ex: Write the expression as the sine of an angle. Then give the exact value.
sin cos cos sin4 12 4 12
sin cos cos sin sin4 12 4 12 4 12
sin sin cos cos sinu v u v u v
1sin
6 2
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.
Ex1: Find the exact value of cos195 .
1. 195 150 45 cos195 cos 150 45
2. cos cos cos sin sinu v u v u v cos 150 45 cos150 cos45 sin150 sin 45
3. 3 2 1 2 6 2 6 2
cos150 cos45 sin150 sin 452 2 2 2 4 4 4
sin sin cos cos sinu v u v u v
cos cos cos sin cosu v u v u v
tan tantan
1 tan tan
u vu v
u v
or
Precalculus Notes: Unit 5 – Trigonometric Identities
Page 10 of 23 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
Ex2: Write as one trig function and find an exact value. tan80 tan55
1 tan80 tan55
1. tan tantan
1 tan tan
u vu v
u v
tan80 tan55
tan 80 551 tan80 tan55
2. tan 80 55 tan 135
3. tan 135 1
Evaluating Trig Functions Given Other Trig Function(s)
Ex: Find cos u v given 15
cos17
u , 3
2u
and 4
sin , 05 2
v v
.
cos cos cos sin sinu v u v u v We must find cosv and sinu .
Draw the appropriate right triangles in the coordinate plane.
15cos
17u ,
3
2u
: 4
sin , 05 2
v v
:
Use the Pythagorean Theorem to find the missing sides.
2 2 215 17
8
U
U
2 2 24 5
3
V
V
In Quadrant III, sine is negative, so 8
sin17
u . In Quadrant I, cosine is positive, so 3
cos5
v .
cos cos cos sin sinu v u v u v 15 3 8 4 45 32 77
17 5 17 5 85 85 85
Proving Identities
Ex: Verify the identity. sin
tan tancos cos
x
y
5 4
v
x
y
15
17
u
Precalculus Notes: Unit 5 – Trigonometric Identities
Page 11 of 23 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
Start with the left side.
sin
tan tancos cos
sin sin cos cos sinu v u v u v
Trig substitution: sin cos cos sin
cos cos
Split the fraction: sin cos cos cos
cos
sin
cos
cos
Simplify: sin sin
cos cos
Trig substitution: tan tan tan tan
You Try: Verify the cofunction identity sin cos2
using the angle difference identity.
QOD: Give an example of a function for which f a b f a f b for all real numbers a and b.
Then give an example of a function for which f a b f a f b for all real numbers a and b.
Precalculus Notes: Unit 5 – Trigonometric Identities
Page 12 of 23 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
Syllabus Objective: 3.3 – The student will simplify trigonometric expressions and prove trigonometric identities (double angle and power-reducing identities).
Ex: Derive the double angle identities using the sum identities.
1. sin 2 sinu u u
sin sin cos sin cos 2sin cosu u u u u u u u
2. cos 2 cosu u u
2 2cos cos cos sin sin cos sinu u u u u u u u
3. tan 2 tanu u u
2
tan tan 2 tantan
1 tan tan 1 tan
u u uu u
u u u
Double Angle Identities
There are two other ways to write the double angle identity for cosine. Use the Pythagorean identity.
2 2
2 2
2 2
sin cos 1
sin 1 cos
cos 1 sin
2 2
2
2
cos2 cos sin
cos2 1 2sin
cos2 2cos 1
Evaluating Double-Angle Trigonometric Functions
Ex: Find the exact value of cos2u given 3
cot 5, 22
u u .
u will be in Quadrant IV and forms a right triangle as labeled.
x
y
u 5
1
2 2
2
sin 2 2sin cos
cos2 cos sin
2tantan 2
1 tan
Precalculus Notes: Unit 5 – Trigonometric Identities
Page 13 of 23 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
Using the Pythagorean Theorem, we have 2 2 25 1 26c c
Double Angle Identity: 2 2cos2 cos sinu u u 5 1
cos , sin26 26
u u
2 2
5 1 25 1 24 12cos2
26 26 26 1326 26u
Note: If u is in Quadrant IV, 3
22
u , then for 2u we have
32 2 2 2 3 2 4
2u u
, which is in Quadrant I. So it makes sense that cos2u is positive.
Solving Trigonometric Equations
Ex: Find the solutions to 4sin cos 1x x in 0,2 .
Rewrite the equation. 2 2sin cos 1x x
Trig substitution. 2sin 2 1x sin 2 2sin cosx x x
Isolate trig function. 1
sin 22
x
Solve for the argument. 12 sin 1x
Because the argument is 2x, we must revisit the domain. 0,2 is the restriction for x. So
0 2x . Therefore, 2 0 2 2 2 0 2 4x x .
5 13 17
2 , 2 , 2 , 26 6 6 6
x x x x
Solve for x. 5 13 17
, , ,12 12 12 12
x
Rewriting a Multiple Angle Trig Function to a Single Angle Ex: Express sin3x in terms of sin x .
Rewrite argument as a sum sin3 sin 2x x x
Sum identity sin cos2 sin 2 cosx x x x
Double angle identities 2
3 2
sin 1 2sin 2sin cos cos
sin 2sin 2sin cos
x x x x x
x x x x
Pythagorean identity 3 2
3 3
sin 2sin 2sin 1 sin
sin 2sin 2sin 2sin
x x x x
x x x x
Simplify 33sin 4sinx x
Precalculus Notes: Unit 5 – Trigonometric Identities
Page 14 of 23 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
Verifying a Trig Identity
Ex: Verify 2
2 tansin 2
1 tan
.
Start with left side. 2
2 tansin 2
1 tan
Pythagorean identity 2
2 tan
sec
Rewrite in sines/cosines
2
sin2
cos1
cos
Simplify 2sin cos
2cos 1
Double angle identity sin 2 2sin cos
Recall:
2 2
2
2
cos2 cos sin
cos2 1 2sin
cos2 2cos 1
Solving for 2sin and 2cos , we can derive the power reducing identities.
2
2
cos2 1 2sin
1 cos2sin
2
2
2
cos2 2cos 1
1 cos2cos
2
22
2
1 cos2sin 1 cos22tan
1 cos2 1 cos2cos2
Power Reducing Identities
2
2
2
1 cos2sin
21 cos2
cos2
1 cos2tan
1 cos2
Ex: Express 5cos x in terms of trig functions with no power greater than 1.
Rewrite as a product 5 2 2cos cos cos cosx x x x
Power reducing identity 1 cos2 1 cos2
cos2 2
x xx
Precalculus Notes: Unit 5 – Trigonometric Identities
Page 15 of 23 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
Multiply 211 2cos2 cos 2 cos
4x x x
Power reducing identity 1 1 cos4
1 2cos2 cos4 2
xx x
You Try:
1. Find the solutions to 2cos sin 2 0x x in 0,2 .
2. Verify 2cos2
cot tansin 2
.
QOD: How do you convert from a cosine function to a sine function? Explain.
Precalculus Notes: Unit 5 – Trigonometric Identities
Page 16 of 23 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
Syllabus Objective: 3.3 – The student will simplify trigonometric expressions and prove trigonometric identities (half angle identities).
Recall: 2 1 cos2sin
2
Let
2
u . We have 2 1 cossin
2 2
u u
Solving for sin2
u
, we have 1 cos
sin2 2
u u
. All of the other half-angle identities can be derived
in a similar manner. Half-Angle Identities
1 cossin
2 2
1 coscos
2 2
1 costan
2 1 cos
u u
u u
u u
u
Note: There are 2 others for tangent.
1 costan
2 sinsin
tan2 1 cos
u u
uu u
u
Note: The will be decided based upon which quadrant 2
u lies in.
Evaluating Trig Functions
Ex: Find the exact value of cos12
.
Rewrite as a half angle 1
cos cos12 2 6
Half angle identity 1 cos
62
Evaluate
31
22
2 3
4
Choose sign 2 3
2
12
is in Quadrant I, where cosine is positive.
Solving a Trig Equation
Ex: Solve the equation sin sin2
xx in 0,2 .
Precalculus Notes: Unit 5 – Trigonometric Identities
Page 17 of 23 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
Half-angle identity 1 cos
sin2
xx
Square both sides 2 1 cossin
2
xx
Pythagorean identity 2 1 cos1 cos
2
xx
Set equal to zero 2
2
2 2cos 1 cos
2cos cos 1 0
x x
x x
Factor 2cos 1 cos 1 0x x
Zero product property
2cos 1 0
1cos
22 4
,3 3
x
x
x
cos 1 0
cos 1
0
x
x
x
2 40, ,
3 3x
You Try:
1. Find the exact value of tan 22.5 .
2. Solve the equation 2sin cos 02
xx in 0,2 .
QOD: Explain why two of the half-angle identities do not have +/− signs.
Precalculus Notes: Unit 5 – Trigonometric Identities
Page 18 of 23 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
Syllabus Objective: 3.5 – The student will solve application problems involving triangles (Law of Sines). Deriving the Law of Sines: Consider the two triangles.
In the acute triangle, sinh
Ab
and sinh
Ba
.
In the obtuse triangle, sin sinh
B Ba
.
Solve for h. sinh b A and sinh a B
Substitute. sin sina B b A can be rewritten as sin sinB A
b a
The same type of argument can be used to show that sin sin sinC B A
c b 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.
sin sin sin
or sin sin sin
A B C a b c
a b 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.
Ex: Solve the triangle ABC given that 15 , 52 , and 9B C b .
We are given AAS, so we will use the Law of Sines.
sin sin sin15 sin52
9
B C
b c c
Solve for c: 9sin52
sin15 9sin52 27.4sin15
c c
Find m A using the triangle sum. 180 15 52 113m A
sin sin sin15 sin113
9
B A
b a a
Solve for a: 9sin113
sin15 9sin113 32sin15
a c
A
a
B
C
9
c
A
a
B
C
b
c
A
a h
B
C
b
c
h a b
c A B
C
Precalculus Notes: Unit 5 – Trigonometric Identities
Page 19 of 23 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
: 113 , 15 , 52 , 32, 9, 27.4ABC m A m B m C a b c
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 54 , 10, 7m C a c .
Given SSA, use Law of Sines. sin sin sin54 sin
7 10
C A A
c a
Solve for A. 110sin54 10sin5410sin54 7sin sin sin
7 7A A A
There is no possible triangle with the given information.
Ex2: Solve the triangle ABC (if possible) when 31 , 46, 29m C b c .
Given SSA, use Law of Sines. sin sin sin31 sin
29 46
C B B
c b
Solve for B.
146sin31 46sin3146sin31 29sin sin sin 54.8
29 29B B B
Note that the calculator only gives the acute angle measure for B. There does exist an obtuse angle B
with the same sine. 180 54.8 125.2m B 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
180 54.8 31 94.2m A 180 125.2 31 23.8m A
sin sin sin31 sin94.2
29
sin31 29sin94.2 56.2
C A
c a a
a a
sin sin sin31 sin 23.8
29
sin31 29sin 23.8 22.7
C A
c a a
a a
Triangle 1: 94.2 , 54.8 , 31 , 56.2, 46, 29m A m B m C a b c
Triangle 2: 23.8 , 125.2 , 31 , 22.7, 46, 29m A m B m C a b c
Application Problems
1. Draw a picture! 2. Use the Law of Sines to solve for what is asked in the problem.
Precalculus Notes: Unit 5 – Trigonometric Identities
Page 20 of 23 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
Ex: 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.
(not drawn to scale)
First, find θ: 180 9 171 . Therefore, the third angle in the small triangle = 5.5°.
Using the law of sines, we know that 13
sin5.5 13sin171 21.22sin5.5 sin171
zz z .
Now we can use right triangle trig to find h: sin3.5 1.3miles21.22
hh
You Try: Solve the triangle ABC (if possible) when 98 , 10, 3m B b c .
QOD: Explain why SSA is the ambiguous case when solving triangles.
3.5° 9°
h
13
z
θ
Precalculus Notes: Unit 5 – Trigonometric Identities
Page 21 of 23 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
Syllabus Objective: 3.5 – The student will solve application problems involving triangles (Law of Cosines). Law of Cosines: For any triangle, ABC
2 2 2
2 2 2
2 2 2
2 cos
2 cos
2 cos
c a b ab C
b a c ac B
a b c bc A
Note: In a right triangle, 2 2 2 2 2 2 2 2 22 cos90 2 0c a b ab c a b ab c a b (Pythagorean
Theorem) The Law of Cosines can be used to solve triangles when given SAS or SSS.
Ex: Solve the triangle ABC when 49 , 42, & 15m A b c .
Note: The given information is SAS. Use 2 2 2 2 cosa b c bc A .
2 2 2 242 15 2 42 15 cos49 1162.36 34.09a a a
Now that we have a matching pair of a side and angle, we can use the Law of Sines.
34.09 42 42sin 49
sin 68.4sin sin sin 34.09sin 49
a bB B
A B B
or 180 68.4 111.6B
Now find the two possibilities for m C using the triangle sum:
180 49 68.4 62.6C or 180 49 111.6 19.4C
Since c is the shortest side, it must be opposite the smallest angle. So 19.4m C .
49 , 111.6 , 19.4 , 34.09, 42, 15m A m B m C a b c
Ex: Solve the triangle ABC when 31, 52, & 28a b c .
Note: The given information is SSS. Use 2 2 2 2 cosa b c bc A . (Or any of them!)
2 2 2 252731 52 28 2 52 28 cos cos 29.8
2912A A A
Now that we have a matching pair of a side and angle, we can use the Law of Sines.
31 52 52sin 29.8sin 56.5
sin sin sin 31sin 29.8
a bB B
A B B
or 180 56.5 123.5B
Now find the two possibilities for m C using the triangle sum:
180 29.8 56.5 93.7C or 180 29.8 123.5 26.7C
Since c is the shortest side, it must be opposite the smallest angle. So 26.7m C .
29.8 , 123.5 , 26.7 , 31, 52, 28m A m B m C a b c
A
a
B
C
b
c
Precalculus Notes: Unit 5 – Trigonometric Identities
Page 22 of 23 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
Application Problems 1. Draw a picture! 2. Use the Law of Cosines to solve for what is asked in the problem.
Ex: A plane takes off and travels 60 miles, then turns 15° and travels for 80 miles. How far is the plane from the airport?
(not drawn to scale) Using the picture, we can find the angle in the triangle to give us SAS.
Use the Law of Cosines: 2 2 2 2 cosc a b ab C
2 2 2 260 80 2 60 80 cos165 19272.89 138.8milesc c c
Area of a Triangle
Recall: 1
2A bh sin sin
hh c
c ;
1sin
2A bc
Formula for the Area of a Triangle Given SAS: 1
sin2
A ab C
Ex: Find the area of triangle ABC shown.
We will use 1
sin2
A ac B .
1 332 50 sin120 16 50 400 3sq units
2 2A A A
Heron’s Area Formula
Semi-Perimeter: 2
a b cs
Area of a Triangle Given SSS: A s s a s b s c
32 50 120°
A
B
C
h
bθ
c h
b
15° 165°
80
60 c
Precalculus Notes: Unit 5 – Trigonometric Identities
Page 23 of 23 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4
Ex: Find the area of the triangle with side lengths 5 m, 6 m, and 9 m.
Semiperimeter: 5 6 9
102 2
a b cs s s
A s s a s b s c
210 10 5 10 6 10 9 10 5 4 1 200 10 2 mA A A A
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.