lesson 5

LESSON 5: TRIGONOMETRIC IDENTITIES

An identity is a relationship that is true for every value of its variable(s) for which each side of the equation is defined.

5.1 FUNDAMENTAL IDENTITIES

Reciprocal identities Quotient Identities

csc θ =

1sin θ tan θ =

sin θcosθ

sec θ =

1cosθ cot θ =

cosθsinθ

cot θ =

1tanθ

Pythagorean Identities Negative Identities

sin²θ + cos²θ = 1 sin(−θ ) = - sin(−θ )

1 + tan²θ = sec²θ cos(−θ ) = cos(−θ ) 1 + cot²θ = csc²θ tan(−θ ) = - tan(−θ )

Proving Identities can be done in two ways:

1. Transforming only one side – making it similar to the other side.2. Proving both sides to the same expression.

No general method of proving an identity can be given. A few suggestions, however, will be of great help.

1. Proving only one side requires the choice of the more complicated side to verify. Have in mind what is to derive in the end.

2. Simplifying of algebraic expressions like factoring, special products, operating fractions, radicals, etc., and applying fundamental identities to make all terms simple in form. If possible, transform the functions to sines and cosines.

4. Correct practice.

Remarks: In proving identities, do not 1. cross multiply2. transpose terms

Note that identities are not required to solve for its variable(s).

Example 1: Prove the following identities.

1. cot θcsc θ

= sin (90°- θ)

Proving the left side:

cos θsinθ1sinθ

?¿ sin (90°- θ)( cosθsin θ )sin θ ?¿ sin (90°- θ) cos θ ?¿ sin (90°- θ)

Let θ be an acute angle in a right triangle, 90° - θ is the complement of θ.

By taking the cofunction of the complementary angles,

sin (90°- θ) = sin (90°- θ)2.

Proving the left side,

As a difference of two squares:

Since, ,

Simplifying,

3.

Adding the functions in the left side,

?

?

?

?

Simplifying the numerator,

cos2 x+(1+2 sinx+sin2 x)cos x¿¿¿

=

Since , ,

2

4.

Replace the left side by identities:

=

5.

Neither the left nor the right side may be called more complicated, therefore, choosing any of the two sides to verify will give the same result. Proving the left side, multiply the conjugate of any of the numerator or denominator, thus, if the conjugate of the numerator is multiplied,

?

?

?

?

?

∙

?

?

6.

Proving the left side, express each function in terms of sine and cosine.

The above result has the same case in no. 5.For variation, if the conjugate of the denominator is multiplied,

?

?

?

7.

Proving the left side, factor the numerator.

Since , ,

8.

Proving the right side, get the common factor.

Since, , use

sec4 B + sec4 B

?

?

?

∙

?( )

9.

Proving the left side, expand each term.

=

By common factoring,

(

10.

Proving the left side, express the tangent function in terms of sine and cosine.

?

?

?

?

?

?

?

?

)

4.2 ADDITION and SUBTRACTION FORMULAS

Addition Formulas

sin(A + B) = sinA cosB + cosA sinBcos(A + B) = cosA cosB – sinA sinB

tan(A+B)=tan A+ tanB1−tan A tan B

Subtraction Formulas

sin(A – B) = sinA cosB – cosA sinBcos(A – B) = cosA cosB + sinA sinB

tan(A−B )=tan A−tanB1+ tan A tan B

Example 2: Find the exact value of the functions using the addition or subtraction formulas.

a. sin 375

Solution:

sin 375 = sin(315+60) = sin315 cos60 + cos315 sin60

= −√22

⋅12+ √22

⋅√32 =

−√24

+ √64

sin 375 =14

(√6−√2 )

b. cos 195

Solution:

cos 195 = cos(150 + 45) = cos150 cos45 – sin150 sin45

= −√32

⋅√22

−12⋅√22

cos 195 =−√2

4(√3+1 )

c. tan 15

Solution:

tan15∘=tan(45∘−30∘)=tan 45∘−tan 30∘

1+tan 45∘ tan 30∘

=1− 1

√3

1+1( 1√3 )=√3−1

√3+1

tan15∘=2−√3

d. sin12 cos33 + cos12 sin33

Solution:

sin12 cos33 + cos12 sin33 = sin(12 + 33) = sin45 =

√22

e. cos15 cos75 – sin15 sin75

Solution:cos15 cos75 – sin15 sin75 = cos(15 + 75) = cos90 = 0

f.

tan15∘−tan 75∘

1+ tan 15∘ tan75∘

Solution:tan15∘−tan 75∘

1+ tan 15∘ tan75∘= tan(15 - 75) = tan(-60) = −√3

Example 3: Given that sin A=3

5 , cosB=− 5

13 , A is in Q2 , B is in Q3 , find the exact value of the functions using the addition or subtraction formulas:

a. sin(A + B)

Solution:

sin (A + B) = sinA cosB + cosA sinB

=

35 (− 5

13 )+(− 45 )(−1213 ) =

3365

b. cos(A – B)

Solution:

cos (A – B) = cosA cosB + sinA sinB

= (−45 )(− 5

13 )+ 35 (−1213 ) =

−1665

c. tan(A + B)

Solution:

tan(A+B)=tan A+ tanB1−tan A tan B

=

− 34+ 125

1−(− 34 )( 125 ) =

3356

Example 4: Prove the identities.

a. sin 3 X=3sin X−4sin3 X

sin(2X+X) =?

sin2X cosX + cos2X sinX =?

2sinXcosX (cosX) + (1 – 2sin2X)sinX =?

2sinXcos2X + sinX – 2sin3X =?

2sinX(1 – sin2X) + sinX – 2sin3X =?

3sin X−4sin3X = 3sin X−4sin3X

b. sin (30∘+θ )+cos (60∘+θ )=cos θ

sin30 cosθ + cos30 sinθ + cos60 cosθ – sin60 sinθ =?

cosθ

12 cosθ +

√32 sinθ +

12 cosθ –

√32 sinθ =

?

( 12+ 12 )cosθ =?

cosθ = cosθ

c.tan(X+ 5

4π )= 1+ tan X1−tan X

tan X+tan 5 π4

1−tan X tan 5 π4

=?

tan X+11− tan X

=?

1+tan X1−tan X

=1+ tan X1−tan X

4.3 DOUBLE-ANGLE FORMULAS

sin 2 A=2sin A cos Acos2 A=cos2A−sin2A

=2cos2A−1

=1−2sin2A

tan2 A= 2 tan A

1−tan2A for tanA ¿ 1

Example 5: Find the exact value of the functions using the double-angle formulas.

a. if tanθ=−1

5, θ in Q2 , find sin 2θ

Solution:

Apply sin 2θ=2sinθ cosθ

=2( 1√26 )(− 5

√26 )=− 513

b. if sinθ=−1

2 , 270 < θ < 360 , find cos2θSolution:

Apply cos2θ=1−2sin2θ

=1−2(−12 )

2

=

12

c. if sin θ=3

5 , θ in Q1 , find tan2θ

Solution:

Apply tan2θ= 2 tan θ

1−tan2θ

=2( 34 )1−( 34 )2 =

=32

1−( 916 ) =

32716 =

247


a.

1−cos2 Xsin 2 X

= tan X

1−(1−2sin2X )2sin X cos X

=?

tan X

1−1+2sin2X2sin X cosX

=?

tan X

sin Xcos X

=?

tan X

tan X=tan X

b. cos 4 A=8cos4 A−8cos 2 A+1

cos2 (2 A ) =?

2cos2 (2 A )−1=?

2 [cos2 A ]2−1=?

2 [2cos2 A−1 ]2−1=?

2 [4 cos4 A−4cos2 A+1 ]−1=?

8cos4 A−8cos 2 A+1=8cos4 A−8cos 2 A+1

c.tan 4 B=

? 4 tanB (1− tan2B )1−6 tan2B+ tan4B

tan2 (2B ) =?

2 tan (2 B )1−tan2 (2B )

=?

2( 2 tanB

1−tan2B )1−[ tan (2 B ) ]2

=?

4 tanB

1−tan2B

1−[ 2 tanB1− tan2B ]2 =?

4 tanB

1−tan2 B

1−[ 4 tan2B

1−2 tan 2B+ tan4 B ]=?

4 tan B

1−tan2 B

1−2 tan2B+ tan4 B−4 tan2B(1−tan2B )2

=?

4 tan B⋅(1− tan2B )

1−6 tan2B+tan4B=?

4 tan B (1− tan2B )1−6 tan2B+ tan4B

=4 tanB (1−tan2B )1−6 tan2B+ tan4B

4.4 HALF-ANGLE FORMULAS

sinθ2=±√ 1−cosθ2

cosθ2=±√ 1+cosθ2

tanθ2=±√ 1−cosθ1+cosθ for cosθ ¿ -1

=1−cosθsinθ for sinθ ¿ 0

=sinθ1+cosθ for cosθ ¿ -1

Example 7: Find the exact value of the functions using the half-angle formulas.

a. sin 292.5º

Solution:

Let

θ2=292.5 °

∴θ=585 °

By sinθ2=√ 1−cosθ2 and since 292.5º is in Q4, sin 292.5º is negative.

Thus, apply

sinθ2=−√ 1−cosθ2 =

−√ 1−cos585 °2 = −√ 1−cos225 °2

= −√ 1+ 1

√22 =

−12√2+√2

b. find cos

θ2 if

sin θ= 513, θ in Q2

Solution:

Since θ is in Q2 and θ when divided by 2 is in Q1,

cosθ2 is positive.

Thus, apply

cosθ2=√ 1+cosθ2 = √ 1−

1213

2 = √ 126 or

√2626

c. find tanθ2 if

cosθ=37 , θ in Q4

Solution:

Since θ is in Q4 and

θ2 when divided by 2 is in Q2,

tanθ2 is negative.

Thus, apply

tanθ2=1−cos θsinθ =

1− 37

−2√107 =

−√105


a.(sin θ2−cos θ2 )

2

=1−sinθ

(√ 1−cosθ2−√ 1+cosθ2 )

2

=?

1−cosθ2

−2√ 1−cosθ2 √ 1+cosθ2+ 1+cosθ

2=?

1−cosθ+1+cosθ2

−2√ 1−cos2θ4=?

22−2√sin2θ4 =

?

1−2(sin θ2 )=?

1−sinθ=1−sin θ

b. tan(B2 +45∘)−tanB=secBtanB2

+ tan 45∘

1−tanB2tan 45∘

−tanB=?

tanB2

+1

1−tanB2

−tanB=?

1−cosBsin B

+1

1− 1−cosBsinB

−tanB=?

1−cosB+sinBsinB

sin B−1+cosBsinB

−tanB=?

sin B−cosB+1sin B+cosB−1

−sin BcosB

=?

cosB (sinB−cosB+1 )−sinB (sin B+cosB−1 )cosB (sinB+cosB−1 )

=?

sin B cosB−cos2B+cosB−sin2B−sinB cosB+sinBcosB (sinB+cosB−1 )

=?

−cos2B−sin2B+sin B+cosBcosB (sinB+cosB−1 )

=?

sin B+cosB−1cosB (sinB+cosB−1 )

=?

1cosB

=?

secB = secB

c. tanX2

+2sin2 X2cot X=sin X

1−cos Xsin X

+2(√ 1−cos X2 )2

(cos Xsin X )=?

1−cos Xsin X

+2( 1−cosX2 )(cosXsin X )=?

(1−cos X ) [1+cos X ]sin X

=?

1−cos2Xsin X

=?

sin2 Xsin X

=?

sin X=sin X

4.5 PRODUCT FORMULAS

sin X cosY=12

[sin (X+Y )+sin (X−Y ) ]cos X sinY=1

2[sin (X+Y )−sin (X−Y ) ]

cos X cosY= 12

[cos (X+Y )+cos (X−Y ) ]sin X sinY=−1

2[cos (X+Y )−cos ( X−Y ) ]

4.6 SUM/DIFFERENCE FORMULASsin A+sin B=2sin 1

2( A+B )cos 1

2(A−B )

sin A−sinB=2cos 12

( A+B ) sin 12

( A−B )

cos A+cosB=2cos 12

(A+B )cos 12

(A−B )

cos A−cosB=−2sin 12

(A+B )sin 12

(A−B )

Example 9: Express each of the following as a sum or difference.

a. sin28ºcos46º=12

[sin (28°+ 46 ° )+sin (28 °−46 ° ) ] =

12

[sin 74 °−sin 18 ° ]

b. cos50ºsin12º=12

[sin (50°+12 ° )−sin (50 °−12 ° ) ]

=12

[sin 62 °−sin 38 ° ]

c. cos63ºcos8º=12

[cos (63°+8 ° )+cos (63 °−8° ) ]

=12

[cos71°+ cos55 ° ]

d. sin125ºsin74º=− 12

[cos (125°+74 ° )−cos (125 °−74 ° ) ]=− 1

2[cos199 °−cos51 ° ]= 1

2[cos51°−cos199 ° ]

Example 10: Express each of the following as a product.

a. sin95º + sin25º=2sin ( 95°+ 25 °2 )cos ( 95 °−25°2 ) =2sin 60 ° cos35 °b. sin63º – sin27º=2cos (

63°+27 °2 ) sin ( 63 °−27 °2 )=2cos 45 °sin 18 °

c. cos48º + cos42º=2cos (48°+ 42°

2 )cos ( 48 °−42°2 )=2cos 45 °cos 3°d. cos78º – cos18º=−2sin ( 78 °+ 18 °2 )sin ( 78 °−18 °2 )=−2sin 48° sin 30 °


a.

sin 5θ+sin 3θcos5θ+cos3θ

=tan 4θ

2sin 12

(5θ+3θ ) cos 12

(5θ−3θ )

2cos 12

(5θ+3θ ) cos 12

(5θ−3θ )=?

sin 4 θcos 4θ

=?

tan 4θ=tan 4θ

b. sin (45∘+A )−sin (45∘−A )=√2sin A2cos 1

2(45°+A+45 °−A ) sin 1

2(45°+ A−45°+ A ) =

?

2cos 45° sin A=?

2(√22 )sin A=?

√2sin A=√2sin A

c. cos220∘+cos100∘+cos20∘=0

(cos220∘+cos100∘ )+cos20∘=?

0

[2cos 12 (220°+100° )cos 12

(220°−100 ° ) ]+cos20 °=?

[2cos160 °cos 60° ]+cos 20°=?

Since 160º is in Q2 and its reference angle is 20º, cos 160º = -cos 20ºThus,

2 (−cos20 ° )( 12 )+cos20 °=?

−cos20°+ cos20 °=?

0 = 0

d. sin 6 X+sin 4 X+sin 2X=4 sin 3 X cos2 X cosX

(sin 6 X+sin 4 X )+sin 2X=?

2sin ( 6 X+ 4X2 )cos ( 6 X−4 X

2 )+sin 2 X=?

2sin 5 X cosX+sin 2 X=?

2sin 5 X cosX+2sin X cos X=?

2cosX (sin 5 X+sin X ) =?

2cosX [2sin ( 5X+X2 )cos ( 5X−X

2 )]=?

4 cosX sin 3 X cos2 X=?

4 sin 3 X cos2 X cosX=4sin 3 X cos 2 X cos X

EXERCISES

Lesson 4.2: Trigonometric Identities

Name: __________________________________ Score: ____________Course/Section: ___________________________ Date: _____________

I. Using the addition and subtraction formulas,

1. find the exact values of sine, cosine and tangent ofa. 165º

Solution:i. sin 165º = sin (135º + 30º) = sin135º cos 30º + cos 135º sin 30º

=

√22 (√32 )+(−√2

2 )( 12 ) =

14

(√6−√2 )

ii. cos 165º = cos (135º + 30º) = cos135º cos 30º - sin 135º sin 30º

= (−√2

2 )(√32 )−√22 ( 12 ) =

−14

(√6+√2 )

iii. tan 165º = tan (135º + 30º)

=

tan135°+ tan 30°1−tan135 ° tan 30 ° =

−1+ √33

1−(−1 )(√33 ) = √3−2

b. 255º

Solution:i. sin 255º = sin (210º + 45º) = sin 210º cos 45º + cos 210º sin 45º

= −12 (√22 )+(−√3

2 )(√22 ) =

−14

(√2+√6 )

ii. cos 255º = cos (210º + 45º) = cos 210º cos 45º - sin 210º sin 45º

= (−√3

2 )(√22 )−(−12 )(√22 ) =

14

(√2−√6 )

iii. tan 255º = tan (210º + 45º)

=

tan210°+ tan 45 °1−tan210 ° tan 45° =

√33

+1

1−√33

(1 ) = 2+√3

c. 375º

Solution:

i. sin 375 = sin(315 + 60) = sin 315 cos 60 + cos 315 sin 60

= −√22 ( 12 )+ √2

2 (√32 ) =

14

(√6−√2 )

ii. cos 375º = cos (315 + 60) = cos 315º cos 60º - sin 315º sin 60º

=

√22 ( 12 )−(−√2

2 )(√32 ) =

14

(√2+√6 )

iii. tan 375º = tan (315 + 60)

=

tan315°+ tan 60 °1−tan315 ° tan 60 ° =

−1+√31−(−1 ) √3 = 2−√3

2. find the exact values of sin ( A+B ) , cos (A+B ) , and tan (A+B ) , given:

a. sin A= 8

17 , tanB= 5

12 , A and B in Q1

Solution:i. sin (A + B) = sinA cosB + cosA sinB

=

817 (1213 )+1517 ( 513 )

=

171221

ii. cos(A + B) = cosA cosB – sinA sinB

=

1517 (1213 )− 8

17 ( 513 ) =

140221

iii. tan(A+B)=tan A+ tanB

1−tan A tan B

=

815

+ 512

1−815 ( 512 )

=

171140

b. cos A=−12

13 , cot B=24

7 , A in Q2 , B in Q3


=

513 (−2425 )+(−1213 )(− 7

25 ) =

−36325


= (−1213 )(−2425 )− 5

13 (− 725 )

=

323325


1−tan A tan B

=

− 512

+ 724

1−(− 512 )( 724 )

= −36323

c. sin A=1

3 , sin B=2

5 , A in Q1 , B in Q2


=

13 (−√21

5 )+ 2√23 ( 25 ) =

4 √2−√2115


=

2√23 (−√21

5 )−13 ( 25 ) =

− 215

(1+√42 )


1−tan A tan B

=

√24

−2√2121

1−√24 (−2√2121 )

=

25√2−9√2182

3. find the exact values of sin ( A−B ) , cos (A−B ) , and tan (A−B ) , given:

a. tan A=3

4 , cosB=12

13 , A and B in Q1

Solution:i. sin (A - B) = sinA cosB - cosA sinB

=

35 (1213 )−45 ( 513 )

=

1665

ii. cos(A - B) = cosA cosB + sinA sinB

=

45 (1213 )+ 35 ( 513 )

=

6365

iii. tan(A−B )=tan A−tanB

1+ tan A tan B

=

34− 512

1+34 ( 512 )

=

1663

b. sin A=16

17 , cosB=−7

8 , A in Q2 , B in Q3

Solution:

i. sin (A - B) = sinA cosB - cosA sinB

=

1617 (−78 )−(−√33

17 )(−√158 )

=

−112−3 √55136


= (−√3317 )(−78 )+1617 (−√15

8 ) =

7√33−16√15136


1+ tan A tan B

=

−16√33

−√157

1+(−16√33 )(√157 ) =

−112−3√557√33−16√15

c. cos A=−2

3 , sin B=−5

8 , A in Q3 , B in Q4

Solution:i. sin (A - B) = sinA cosB - cosA sinB

= −√53 (√398 )−(−23 )(−58 )

=

−√195−1024


= (−23 )(√398 )+(−√5

3 )(−58 ) =

5√5−2√3924


1+ tan A tan B

=

√52

−(− 5

√39 )1+ √5

2 (− 5√39 )

=

√195+102√39−5√5

4. find the exact values using the addition and subtraction formulasa. sin210ºcos105º + cos210ºsin105º

b. cos15ºcos75º – sin15ºsin75º

c.

tan 45 °−tan135 °1+ tan 45 ° tan 135 °

5. Prove the identities.

a. tan (45°+A )= 1+ tan A

1−tan A

b.

sin ( A+B )cos A cosB

=tan A+tan B

c. sin A cos ( A−B )−cos A sin (A−B )=sin B

d. cos ( π3−X)=12 cosX+ √3

2sin X

e. sec (X+2π )=sec X

f. sin (X+10 ° ) cos (X−10 ° )−cos ( X+10 ° )sin (X−10 ° )=sin 20 °

II. Using the double-angle formulas,

1. find the exact values of sin2θ , cos2θ , and tan2θ , given:

a. cot θ=−4

3 , 90°<θ<180°

Solution:

i. Apply sin 2θ=2sinθ cosθ

= 2( 35 )(−45 )

= −2425

ii. Apply cos2θ=1−2sin2θ

= 1−2( 35 )

2

=

725

iii. Apply tan2θ= 2 tan θ

1−tan2θ

=

2(− 34 )1−(−34 )

2

=

−64

1−( 916 ) =

−32716 =

−247

b. sin θ=1

3 , θ in Q1

Solution:


= 2( 13 )( 2√23 )

=

4 √29

ii. Apply cos2θ=1−2sin2θ

= 1−2( 13 )

2

=

79


1−tan2θ

=

2(√24 )1−(√24 )

2

=

2√24

1−( 216 ) =

√221416 =

4 √27

c. secθ=25

7 , csc θ=−25

24

Solution:


= 2(−2425 )( 725 )

= −336625

ii. Apply cos2θ=2cos2θ−1

= 2( 725 )

2

−1 =

−527625


1−tan2θ

=

2(−247 )1−(−247 )

2

=

−487

1−(57649 ) =

−487

−52749 =

336527


a. sin 4 A=cos A (4 sin A−8sin3 A )

b. sec 2X=sec

2X2−sec2X

c. tan3 B=3 tan B− tan

3B1−3 tan2B

d.

1+sin 2θcos2θ

=cosθ+sin θcosθ−sinθ

e.

sin 3u+cos3ucosu−sinu

=1+2sin 2u

sin 2ucosu+cos2u sinu+cos 2ucosu−sin 2u sinucosu−sinu =

sin 2u (cosu−sinu )+cos2u (cosu+sinu )cosu−sinu =

sin 2u (cosu−sinu )+ (cos2u−sin2u ) (cosu+sinu )cosu−sinu =

(cosu−sinu ) [ sin2u+(cos u+sinu )2 ]cosu−sinu =

sin 2u+(cos2u+2sin ucosu+sin2u ) = 1+2sin 2u=1+2sin 2u

f. cos4 β−sin4 β=cos2 β

III. Using the half-angle formulas,

1. find the exact values of a. sin 75º

Let

θ2=75 °

∴θ=150 °Since 75º is in Q1, sin 75º is positive.

Thus, apply sinθ2=√ 1−cosθ2

Then, sin 75°=√ 1−cos150 °2 = √ 1−(−√3

2 )2 =

14

(√2+√6 )

b. cos 67.5º

Let

θ2=67 .5 °

∴θ=135 °Since 67.5º is in Q1, cos 67.5º is positive.

Thus, apply cos

θ2=√ 1+cosθ2

Then, cos67 .5°=√ 1+cos 135°2 = √ 1+(−√2

2 )2 =

12√2−√2

c. sinθ2 , if secθ=−7 , 90º < θ < 180º

Since θ is in Q2, sinθ2 is positive.

Thus, apply sinθ2=√ 1−cosθ2

Then, sinθ2 = √ 1−(−17 )

2 =

2√77

d. cos

θ2 , if

tanθ=34 , 180º < θ < 270º

Since θ is in Q3, cos

θ2 is negative.

Thus, apply cos

θ2=−√ 1+cosθ2

Then, cos

θ2 =

−√ 1+(−45 )2 =

−√1010


a. cos A=1−2sin2 A

2 Recall: cos2θ=1−2sin2θ

∴ cos A=1−2sin2 A

2 Let:θ= A

2 then 2θ=A

b. tanB2

=cscB−cotB

=

1sin B

−cosBsinB

=

1−cosBsin B Recall:

tanθ2=1−cos θsinθ

∴ tanB2

=tan B2 Let: θ=B then

θ2=B2

Thus, tanB2

=1−cosBsinB

c.

sinθ=2 tan

θ2

1+ tan2θ2

=

2sin θcos θ1

cos2θ Recall: sin 2 A=2sin A cos A

= 2sin

θ2cos

θ2 Let:

A=θ2 then 2 A=θ

∴ sinθ = sinθ Thus, sin θ=2sin θ

2cosθ2

d.

cos X=1−tan2 X

2

1+ tan2X2

=

1−sin2

X2

cos2X2

1

cos2X2 Recall: cos2θ=cos

2θ−sin2θ

=

cos2X2

−sin2 X2

cos2X2 •

cos2X2 Let:

θ= X2 then 2θ=X

∴ cos X = cos X Thus, cos X = cos2

X2

−sin2 X2

e. 1+ tanU tan U

2=secU

1+[sinUcosU ( 1−cosUsinU )]

=

1+( 1−cosUcosU )

=

cosU+1−cosUcosU =

sec U = sec U

f. (1+cot2 ω2 )(sinω tan ω2 )=2

(csc2 ω2 )(sinωsinω2

cosω2

) =

( 1

sin2ω2 )(sinωsin

ω2

cosω2

) = Recall: sin 2θ=2sinθ cosθ

sinω

sinω2cos

ω2

⋅22

= Let:θ=ω

2 then 2θ=ω

2sinωsinω = Thus,

sinω=2sin ω2cos

ω2

2 = 2

IV. Express each of the following as a sum or difference.

1. sin142º sin21º = −12

[cos (142°+21 ° )−cos (142 °−21 ° ) ]

= −12

[cos163 °−cos121 ° ] =

12

[cos121 °−cos163 ° ]

2. cos43º cos48º =

12

[cos (43°+48 ° )+cos (43°−48 ° ) ]

=

12

[cos91 °+cos (−5° ) ] =

12

[cos91°+cos5 ° ]

3. sin44º cos105º =

12

[sin (44 °+105° )+sin (44 °−105 ° ) ]

=

12

[sin (149 ° )+sin (−61° ) ] =

12

[sin 149°−sin 61° ]

4. cos27º sin46º =

12

[sin (27°+46 ° )−sin (27 °−46 ° ) ]

=

12

[sin 73 °−sin (−19 ° ) ] =

12

[sin 73°+ sin19 ° ]

5. sin123º sin59º = −12

[cos (123°+59 ° )−cos (123°−59° ) ]

= −12

[cos182 °−cos64 ° ] =

12

[cos64 °−cos182 ° ]

6. cos7A sin4A =

12

[sin (7 A+4 A )−sin (7 A−4 A ) ]

=

12

[sin 11A−sin 3 A ]

7. sin54º cos17º =

12

[sin (54°+17 ° )+sin (54 °−17 ° ) ]

=

12

[sin 71°+sin 37 ° ]

8. cos125º cos74º =

12

[cos (125°+74 ° )+cos (125°−74 ° ) ]

=

12

[cos199°+cos51 ° ]

9. sin

A4 cos

A4 =

12

[sin (54°+17 ° )+sin (54 °−17 ° ) ]

=

12

[sin 71°+sin 37 ° ]

10.cos(90 °−5 A ) sin(90 °−2 A ) =

12

[sin (180 °−7 A )−sin (−3 A ) ]

=

12

[sin (180 °−7 A )+sin 3 A ]

=

12

[ (sin 180 °cos 7 A−cos 180° sin 7 A )+sin 3 A ]

=

12

[sin 7 A+sin 3 A ]

V. Express each of the following as a product.

1. sin20º + sin10º = 2sin(20°+10 °2 )cos(20 °−10°2 )

= 2sin 15° cos5 °

=

12

(√6−√2 )cos5 °

2. cos62º – cos24º = −2sin(62°+24 °2 )sin(62 °−24 °2 )

= −2sin 43 ° sin 19°

3. cos40º + cos32º = 2cos(40°+32 °2 )cos (40 °−32°2 )

= 2cos36 ° cos4 °

4. sin28º – sin27º = 2cos(28°+27 °2 )sin(28°−27 °2 )

= 2cos27 .5° sin 0 .5°

5. sin14º + sin67º = 2sin(14 °+67 °2 )cos (14 °−67°2 )

= 2sin 40 .5° cos (−26 .5 ° ) = 2sin 40 .5° cos26 .5 °

6. cos71º – cos8º = −2sin(71°+8 °2 )sin (71°−8 °2 )

= −2sin 39.5 ° sin31 .5°

7. sin67º – sin54º = 2cos(67°+54 °2 )sin (67 °−54 °2 )

= 2cos60 .5° sin 6 .5 °

8. cos38º + cos26º = 2cos(38°+26 °2 )cos (38 °−26 °2 )

= 2cos32 °cos 6 °

9. sin33º – sin56º = 2cos(33°+56 °2 )sin(33°−56 °2 )

= 2cos 44 .5 ° sin (−11.5 ° ) = −2cos 44 .5 ° sin 11.5 °

10.cos28º – cos32º = −2sin(28°+32 °2 )sin (28 °−32 °2 )

= −2sin 30 °sin (−2 ° )

= 2sin 30° sin (−2 ° ) = sin 2º

VI. Prove the identities.

1. sin40º + sin20º = cos10º

2sin(40°+20 °2 )cos (40 °−20°2 )

= 2 sin 30º cos 10º =

2 ( 12 ) cos 10º =

cos 10º = cos 10º

2.

sin 75 °−sin 15 °cos75°+cos 15°

= 1

√3

2cos(75°+15 °2 )sin(75 °−15 °2 )2cos (75°+15 °2 )cos (75 °−15 °2 )

=

2cos45 ° sin 30°2cos 45 ° cos30 ° =

12

√32 =

1

√3 =

1

√3

3.

sinθ+sin 3θcosθ+cos3θ

=tan 2θ

2sin( θ+3θ2 )cos(θ−3θ2 )2cos( θ+3θ2 )cos( θ−3θ2 )

= tan 2θ = tan 2θ

4. cos A+cos2 A+cos3 A=cos2 A (1+2cosA )(cos A + cos 3A) + cos 2A =

2cos( A+3 A2 )cos( A−3 A

2 )+cos2 A =

2cos 2A cos(-A) + cos 2A =

cos2 A (1+2cos A ) = cos2 A (1+2cos A )

5. sin X+sin2 X+sin3 X=sin 2 X (1+2cosX )(sin 3X + sin X) + sin 2X =

2sin( 3 X+X2 )cos( 3 X−X

2 )+sin 2 X =

6. cos130°+cos 110°+cos10°=0

lesson 5

Documents