precalculus notes: unit 5 – trigonometric identities page 1 of 23

Precalculus Notes: Unit 5 – Trigonometric Identities

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 cot

1 1 1csc sec cot

sin cos tan

Quotient Identities sin cos

tan cotcos sin

Recall: Unit Circle 1, cos , sinr x y or cos sin1 1

x yx y

Pythagorean Theorem: 2 2

1x y Pythagorean Identity: 2 2

sin cos 1

To derive the other Pythagorean Identities, divide the entire equation by 2

sin and then by 2

cos :

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

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.

a) 2cot 1 cos

coscot

sin

2 2 2 2sin cos 1 sin 1 cos

,x y

Note:

22sin sin



2 2cos coscot 1 cos sin

sin sinsin sin cos sin

b) tan csc sin

tancos

1

cscsin

sintan csc

1

cos sin

1sec

cos

Ex2: 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 2

1 cot csc cot csc 1

2

2 2 22

2 2 2

cos

csc 1 cot cossin

cos cos cos

x

x x xx

x x x 2 2sin cosx x

2

2

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.

Ex3: Solve the equation 2

4 sin 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 2x

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

a

b

c

θ

Φ



**The trig functions of are equal to the cofunctions of θ, when and are complementary.

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

x

z

z

y

θ −θ

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



Ex4: 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.

Ex5: Simplify the expression by factoring. 3 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 2

sin cos 1x x

Ex6: 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 x

1csc

sinsinx

xx

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 .

Ex7: 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.

Ex8: Solve the equation by extracting square roots. 2

4 sin 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.

Ex9: Solve the equation by factoring. 2

2 cos cos 1x x

Set equal to zero. 2

2 cos 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

2

5,

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.



Ex10: 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.

10

cos x

x

1sin

2

5,

6 6

x

x

Note: It is possible for an equation to have no solution.

Ex11: Solve by rewriting in a single trig function. 2

2 sin 3cos 3x x

Substitute Pyth. Identity. 22 1 cos 3cos 3x x 2 2 2 2

sin cos 1 sin 1 cos

Simplify algebraically. 2 22 2cos 3cos 3 2cos 3cos 1 0x x x x

Factor and solve. 1

2cos 1 cos 1 0 cos , cos 12

x x x x 5

0, ,3 3

x

Ex12: Solve using trig substitutions. 3

sintan

cos

xx

x

Rewrite 3 2

sin 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

Ex13: 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 1

cos4

x 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. 2

4 , 2cosx x

QOD: Explain the relationship between trig functions and their cofunctions.



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 2

sin cos 1 cos 1 sinx x x x

Ex2: 21 1

2sec1 sin 1 sin

xx x

Start with the left side.

Combine fractions.

21 12sec

1 sin 1 sin

1 sin 1 sin

1 sin 1 sin

xx x

x 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 2

2sec 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

Trig substitution. 2 2sec cos 1x x 2 2

tan 1 secx x

Trig substitution. 2 2sec sinx x 2 2 2 2

sin cos 1 cos 1 sinx x x x


2

1sin

cosx

x

1sec

cosx

x

Multiply. 2

2

sin

cos

x

x

Identitiy. 2 2tan tanx x

sintan

cos

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. 2

1 sin

cos 1 sin

x

x x


cos

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

2

2

sec 1sin

sec

Start with left side. 2

2

2

sec 1sin

sec

Split the fraction. 2

2 2

sec 1

sec sec

Simplify. 2

11

sec


1 cos 1

cossec

Identity. 2 2

sin sin 2 2 2 2

sin cos 1 sin 1 cos



Challenge: Try to prove the identity above in another way.

You Try: Prove the identity. 2 2

cos sin cos sin 2x x x x

QOD: List at least 5 strategies you can use when proving trigonometric identities.




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 2

2 3 5 25 2 2 2

2 3 2 3 13

So in general, 2 2 2

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.

Ex1: 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

2 1sin sin

12 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.

Ex2: 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 sin45


cos cos cos sin sinu v u v u v

tan tantan

1 tan tan

u vu v

u v

or



3. 3 2 1 2 6 2 6 2

cos150 cos45 sin150 sin 452 2 2 2 4 4 4

Ex3: Write as one trig function and find an exact value. tan80 tan55

1 tan80 tan55

1. tan tan

tan1 tan tan

u vu v

u v

tan80 tan55tan 80 55

1 tan80 tan55

2. tan 80 55 tan 135

3. tan 135 1

Evaluating Trig Functions Given Other Trig Function(s)

Ex4: Find cos u v given 15

cos17

u , 3

2u and

4sin , 0

5 2v 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 :

4sin , 0

5 2v 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 v15 3 8 4 45 32 77

17 5 17 5 85 85 85

5

4

v

15

17

u



Proving Identities

Ex5: Verify the identity. sin

tan tancos cos

Start with the left side.

sin

tan tancos cos


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.




trigonometric identities (double angle and power-reducing identities).

Ex1: 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

cos 2 cos sin

cos 2 1 2sin

cos 2 2cos 1

2 2

2

sin 2 2sin cos

cos 2 cos sin

2 tantan 2

1 tan



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.

Using the Pythagorean Theorem, we have 2 2 25 1 26c c

Double Angle Identity: 2 2

cos 2 cos sinu u u 5 1

cos , sin26 26

u u

2 25 1 25 1 24 12

cos226 26 26 1326 26

u

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 IV. So it makes sense that cos2u is positive.

Solving Trigonometric Equations

Ex2: Find the solutions to 4sin cos 1x x in 0,2 .

Rewrite the equation. 2 2sin cos 1x x

Trig substitution. 2sin 2 1x sin2 2sin cosx x x

Isolate trig function. 1

sin 22

x

Solve for the argument. 1 1

2 sin2

x

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

u

5

1



Rewriting a Multiple Angle Trig Function to a Single Angle

Ex3: Express sin3x in terms of sin x .

Rewrite argument as a sum sin 3 sin 2x x x

Sum identity sin cos2 sin2 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

Verifying a Trig Identity

Ex4: 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 2

sin cos2

cos 1

Double angle identity sin2 2sin cos

Recall:

2 2

2

2

cos 2 cos sin

cos 2 1 2sin

cos 2 2cos 1

Solving for 2

sin and 2

cos , we can derive the power reducing identities.

2

2

cos 2 1 2 sin

1 cos 2sin

2

2

2

cos 2 2 cos 1

1 cos 2cos

2

22

2

1 cos2

sin 1 cos22tan1 cos2 1 cos2cos

2



Power Reducing Identities

2

2

2

1 cos 2sin

2

1 cos 2cos

2

1 cos 2tan

1 cos 2

Ex5: 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 cos 2 1 cos 2

cos2 2

x xx

Multiply 21

1 2cos2 cos 2 cos4

x x x

Power reducing identity 1 1 cos 4

1 2 cos 2 cos4 2

xx x

You Try:

1. Find the solutions to 2cos sin2 0x x in 0,2 .

2. Verify 2cos 2

cot tansin 2

.

QOD: How do you convert from a cosine function to a sine function? Explain.




trigonometric identities (half angle identities).

Recall: 2 1 cos 2

sin2

Let 2

u. We have

2 1 cossin

2 2

u u

Solving for sin2

u, we have

1 cossin

2 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 sin

sintan

2 1 cos

u u

u

u u

u

Note: The will be decided based upon which quadrant 2

u lies in.

Evaluating Trig Functions

Ex1: Find the exact value of cos12

.

Rewrite as a half angle 1

cos cos12 2 6

Half angle identity 1 cos

6

2

Evaluate

31

2

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 cos12 3 4



Solving a Trig Equation

Ex2: Solve the equation sin sin2

xx in 0,2 .

Half-angle identity 1 cos

sin2

xx

Square both sides 2 1 cos

sin2

xx

Pythagorean identity 2 1 cos

1 cos2

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

2

2 4,

3 3

x

x

x

cos 1 0

cos 1

0

x

x

x

2 4

0, ,3 3

x

You Try:

1. Find the exact value of tan 22.5 .

2. Solve the equation 2

sin cos 02

xx in 0,2 .

QOD: Explain why two of the half-angle identities do not have +/− signs.



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.

Ex1: 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:

9sin52sin15 9sin52 27.4

sin15c 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:

9sin113sin15 9sin113 32

sin15a c

: 113 , 15 , 52 , 32, 9, 27.4ABC m A m B m C a b 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



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.

1 110sin54 10sin5410sin54 7sin sin sin sin 1.1557

7 7A A A

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 31 , 46, 29m C b c .

Given SSA, use Law of Sines. sin sin sin 31 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 sin 31 sin 94.2

29

sin 31 29sin 94.2 56.2

C A

c a a

a a

sin sin sin 31 sin 23.8

29

sin 31 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.

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.

(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

sin 5.5 13sin171 21.22sin 5.5 sin171

zz z .

Now we can use right triangle trig to find h: sin 3.5 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



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.

Ex1: 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 Bor 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

Ex2: 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 2527

31 52 28 2 52 28 cos cos 29.82912

A 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 Bor 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



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?

(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 22 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 3 sq 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

b

θ

c h

b

15°

165°

80

60

c



Ex4: 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.

precalculus notes: unit 5 – trigonometric identities page 1 of 23

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