Chapter 5 Trigonometric Equations€¦ · 11/5/2013 · 5/10/13 Audacity Obj: SWBAT simplify trig...

MATHPOWERTM 12, WESTERN EDITION

5.4

5.4.1

Chapter 5 Trigonometric Equations

5/10/13 Audacity

Obj: SWBAT simplify trig identities Trigonometric identities help you solve equations that arise when studying geometry, and in all kinds of higher

mathematics. I don't know what kind of level your mathematics is at, but rest assured that they will come in handy if

you do any senior or university level mathematics.

Trigonometry comes up in many places because of its links with geometry, examples are architecture, engineering,

even medicine. Trig identities also come up alot in physics and pure mathematics because Trig functions are the

simplest and most natural kinds of waves, and sometimes to analyse these ways we need to transform them into a

different form. Here trig identities come into play. Waves come up absolutely everywhere including physics, pure

mathematics, chemistry, sound engineering, civil engineering, aeronatics, they're just everywhere!

Bell Ringer: Simplify sinx cosx tanx cscx

HW Requests:

Pg 395 #72-75, 79, 80

Homework:

WS 14.3 Simplify Trig Identities

Education is Power!

Dignity without compromise!

A trigonometric equation is an equation that involves

at least one trigonometric function of a variable. The

equation is a trigonometric identity if it is true for all

values of the variable for which both sides of the

equation are defined.

Trigonometric Identities

Prove that tan sin

cos.

y

x

y

r

x

r

y

r

r

x

y

x

L.S. = R.S. 5.4.2

Recall the basic

trig identities:

sin y

r

cos x

r

tan y

x

5.4.3

Basic Trigonometric Identities

Quotient Identities

tan sin

coscot

cos

sin

Reciprocal Identities

sin 1

csccos

1

sectan

1

cot

Pythagorean Identities

sin2 + cos2 = 1 tan2 + 1 = sec2 cot2 + 1 = csc2

sin2 = 1 - cos2

cos2 = 1 - sin2

tan2 = sec2 - 1 cot2 = csc2 - 1

5

Trigonometric identities

• sin2A + cos2A = 1

• 1 + tan2A = sec2A

• 1 + cot2A = cosec2A

• sin(A+B) = sinAcosB + cosAsin B

• cos(A+B) = cosAcosB – sinAsinB

• tan(A+B) = (tanA+tanB)/(1 – tanAtan B)

• sin(A-B) = sinAcosB – cosAsinB

• cos(A-B)=cosAcosB+sinAsinB

• tan(A-B)=(tanA-tanB)(1+tanAtanB)

• sin2A =2sinAcosA

• cos2A=cos2A - sin2A

• tan2A=2tanA/(1-tan2A)

• sin(A/2) = ±{(1-cosA)/2}

• Cos(A/2)= ±{(1+cosA)/2}

• Tan(A/2)= ±{(1-cosA)/(1+cosA)}

Identities can be used to simplify trigonometric expressions.

Simplifying Trigonometric Expressions

cos sin tan

cos sin

sin

cos

cos

sin2

cos

cos 2 sin2

cos

1

cos

sec

a)

Simplify.

b) cot2

1 sin2

cos 2

sin2 cos

2

1

1

sin2

csc2

5.4.5

cos 2

sin2

1

cos2

5.4.6

Simplifing Trigonometric Expressions

c) (1 + tan x)2 - 2 sin x sec x

1 2 tanx tan2x 2

sinx

cosx

1 tan2x 2tanx 2 tanx

sec2x

d) cscx

tan x cot x

1

sinx

sinx

cos x

cosx

sinx

1

sinx

sin2x cos

2x

sinxcos x

1

sinx

sinx cos x

1

cos x

1

sinx

1

sinx cos x

(1 tanx)2

2 sinx1

cosx

End here 5/10

sinA cos A 2

sin2A 2sinAcos A cos

2A

1 2sinAcos A

cos A

sinA1

sinA

cos A

sinA

sinA

1

= cosA

Trigonometric Identities [cont’d] Start with pg 446 Ex 1

5.4.4


** just combined fractions


Squared both sides…


Square root of both sides…

5.4.7

Proving an Identity

Steps in Proving Identities

1. Start with the more complex side of the identity and work

with it exclusively to transform the expression into the

simpler side of the identity.

2. Look for algebraic simplifications:

• Do any multiplying , factoring, or squaring which is

obvious in the expression.

• Reduce two terms to one, either add two terms or

factor so that you may reduce. 3. Look for trigonometric simplifications:

• Look for familiar trig relationships.

• If the expression contains squared terms, think

of the Pythagorean Identities. • Transform each term to sine or cosine, if the

expression cannot be simplified easily using other ratios.

4. Keep the simpler side of the identity in mind.

22 sincoscosecsin Establish the following identity:

In establishing an identity you should NOT move things

from one side of the equal sign to the other. Instead

substitute using identities you know and simplifying on

one side or the other side or both until both sides match.

22 sincoscosecsin

Let's sub in here using reciprocal identity

22 sincossin

1sin

22 sincos1

We often use the Pythagorean Identities solved for either sin2 or cos2.

sin2 + cos2 = 1 solved for sin2 is sin2 = 1 - cos2 which is our

left-hand side so we can substitute.

22 sinsin

We are done!

We've shown the

LHS equals the

RHS

cos1

sincotcosec

Establish the following identity:

Let's sub in here using reciprocal identity and quotient

identity

Another trick if the

denominator is two terms

with one term a 1 and the

other a sine or cosine,

multiply top and bottom of

the fraction by the

conjugate and then you'll

be able to use the

Pythagorean Identity on

the bottom

We worked on

LHS and then

RHS but never

moved things

across the = sign

cos1

sincotcosec

cos1

sin

sin

cos

sin

1

cos1

sin

sin

cos1

combine fractions

cos1

cos1

cos1

sin

sin

cos1

2cos1

cos1sin

sin

cos1

FOIL

denominator

2sin

cos1sin

sin

cos1

sin

cos1

sin

cos1

5.4.8

Proving an Identity

Prove the following:

a) sec x(1 + cos x) = 1 + sec x

= sec x + sec x cos x

= sec x + 1

1 + sec x

L.S. = R.S.

b) sec x = tan x csc x

sinx

cos x

1

sinx

1

cos x

secx

secx

L.S. = R.S.

c) tan x sin x + cos x = sec x

sinx

cos x

sinx

1 cosx

sin2 x cos 2 x

cos x

1

cos x

secx

secx

L.S. = R.S.

d) sin4x - cos4x = 1 - 2cos2 x

= (sin2x - cos2x)(sin2x + cos2x)

= (1 - cos2x - cos2x)

= 1 - 2cos2x

L.S. = R.S.

1 - 2cos2x

e)

1

1 cos x

1

1 cosx 2 csc

2x

(1 cos x) (1 cosx)

(1 cosx)(1 cos x)

2

(1 cos2

x)

2

sin2x

2csc2x

2csc2x

L.S. = R.S.

Proving an Identity

5.4.9

Proving an Identity

5.4.10

f)

cos A

1 sinA

1 sinA

cos A 2 secA

cos 2 A (1 sinA)(1 sinA)

(1 sinA)(cos A)

cos 2 A (1 2sinA sin2 A)

(1 sinA)(cos A)

cos 2 A sin2 A 1 2sinA

(1 sinA)(cos A)

2 2sinA

(1 sinA)(cos A)

2(1 sinA)

(1 sinA)(cos A)

2

(cos A)

2secA

2secA

L.S. = R.S.

Using Exact Values to Prove an Identity

5.4.11

Consider sinx

1 cos x

1 cosx

sinx.

b) Verify that this statement is true for x =

6.

a) Use a graph to verify that the equation is an identity.

c) Use an algebraic approach to prove that the identity is true

in general. State any restrictions.

y 1 cos x

sinxy

sinx

1 cos xa)

sinx

1 cos x

1 cosx

sinx

1

2

1 3

2

b) Verify that this statement is true for x =

6.

sin

6

1 cos

6

1

2

2

2 3

1

2 3

1 cos

6

sin

6

1 3

2

1

2

2 3

2

2

1

2 3

2 3

1

2 3

2 3

2 3

2 3

4 3

2 3

Rationalize the

denominator:

1

2 3

L.S. = R.S.

Using Exact Values to Prove an Identity [cont’d]

5.4.12

Therefore, the identity is

true for the particular

case of x

6.



Using Exact Values to Prove an Identity [cont’d]

5.4.13

sinx

1 cos x

1 cosx

sinx

sinx

1 cos x

1 cos x

1 cos x

sinx(1 cosx)

1 cos2

x

sinx(1 cosx)

sin2x

1 cos x

sinx

1 cos x

sinx

L.S. = R.S.

Note the left side of the

equation has the restriction

1 - cos x ≠ 0 or cos x ≠ 1.

Therefore, x ≠ 0 + 2 n, where n is any integer.

The right side of the


sin x ≠ 0. x = 0 and Therefore, x ≠ 0 + 2n

and x ≠ + 2n, where n is any integer.

Restrictions:

Proving an Equation is an Identity

Consider the equation sin2 A 1

sin2

A sinA 1

1

sinA.

b) Verify that this statement is true for x = 2.4 rad.

a) Use a graph to verify that the equation is an identity.



y sin2 A 1

sin2

A sinAy 1

1

sinA

a)

5.4.14

b) Verify that this statement is true for x = 2.4 rad.

Proving an Equation is an Identity [cont’d]

sin2 A 1

sin2

A sinA 1

1

sinA

(sin 2.4)2 1

(sin 2.4)2

sin2.4

= 2.480 466

1

1

sin 2.4

= 2.480 466

Therefore, the equation is true for x = 2.4 rad.

L.S. = R.S.

5.4.15

5.4.16

Proving an Equation is an Identity [cont’d]

sin2 A 1

sin2

A sinA 1

1

sinA



(sinA 1)(sinA 1)

sinA(sinA 1)

(sinA 1)

sinA

sinA

sinA

1

sinA

1 1

sinA

1 1

sinA

L.S. = R.S.

Note the left side of the

equation has the restriction:

sin2A - sin A ≠ 0

A 0, or A

2

Therefore, A 0 2 n or

A + 2n, or

A

2 2 n, where n is

any integer.

The right side of the


sin A ≠ 0, or A ≠ 0.

Therefore, A ≠ 0, + 2 n, where n is any integer.

sin A(sin A - 1) ≠ 0

sin A ≠ 0 or sin A ≠ 1

Chapter 5 Trigonometric Equations€¦ · 11/5/2013 · 5/10/13 Audacity Obj: SWBAT simplify trig...

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Transcript of Chapter 5 Trigonometric Equations€¦ · 11/5/2013 · 5/10/13 Audacity Obj: SWBAT simplify trig...