M24 Sum, Difference, CoFunctions and Double Identities .pdf

Sum and Difference

Identities,Co-Function, and Double

Measure Identities

Objectives:

1 To illustrate the identities involving sums and differences of real

numbers

2 To demonstrate the co-function identities

β

α

α−β

M(1,0)

P Q

α−β

M(1,0)

R

P Q=RM

P has coordinates (cosα,sinα)

Qhas coordinates (cosβ,sinβ)

β

α

α−β

M(1,0)

P

Q

α−β

M(1,0)

R

P Q=RM

P has coordinates (cosα,sinα)

Qhas coordinates (cosβ,sinβ)

β

α

α−β

M(1,0)

P Q

α−β

M(1,0)

R

P Q=RM

P has coordinates (cosα,sinα)

Qhas coordinates (cosβ,sinβ)

β

α

α−β

M(1,0)

P Q

α−β

M(1,0)

R

P Q=RM

P has coordinates (cosα,sinα)

Qhas coordinates (cosβ,sinβ)

β

α

α−β

M(1,0)

P Q

α−β

M(1,0)

R

P Q=RM

P has coordinates (cosα,sinα)

Qhas coordinates (cosβ,sinβ)

β

α

α−β

M(1,0)

P Q

α−β

M(1,0)

R

P Q=RM

P has coordinates (cosα,sinα)

Qhas coordinates (cosβ,sinβ)

β

α

α−β

M(1,0)

P Q

α−β

M(1,0)

R

P Q=RM

P has coordinates (cosα,sinα)

Qhas coordinates (cosβ,sinβ)

β

α

α−β

M(1,0)

P Q

α−β

M(1,0)

R

P Q=RM

P has coordinates

(cosα,sinα)

Qhas coordinates (cosβ,sinβ)

β

α

α−β

M(1,0)

P Q

α−β

M(1,0)

R

P Q=RM

P has coordinates (cosα,sinα)

Qhas coordinates (cosβ,sinβ)

β

α

α−β

M(1,0)

P Q

α−β

M(1,0)

R

P Q=RM

P has coordinates (cosα,sinα)

Qhas coordinates (cosβ,sinβ)

β

α

α−β

M(1,0)

P Q

α−β

M(1,0)

R

P Q=RM

P has coordinates (cosα,sinα)

Qhas coordinates (cosβ,sinβ)

P Q=RM

By distance formula: p

(cosα−cosβ)2_{+ (sinα−sinβ)}2₌p

(cos(α−β)−1)2_{+ (sin(α−β)−0)}2

Square and expand the LHS:

cos2α−2 cosαcosβ+ cos2β+ sin2α−2 sinαsinβ+ sin2β

= 1 + 1−2(cosαcosβ+ sinαsinβ) = 2−2(cosαcosβ+ sinαsinβ)

Square and expand the RHS:

P Q=RM

By distance formula: p

(cosα−cosβ)2_{+ (sinα−sinβ)}2₌p

(cos(α−β)−1)2_{+ (sin(α−β)−0)}2

Square and expand the LHS:

cos2α−2 cosαcosβ+ cos2β+ sin2α−2 sinαsinβ+ sin2β

= 1 + 1−2(cosαcosβ+ sinαsinβ) = 2−2(cosαcosβ+ sinαsinβ)

Square and expand the RHS:

P Q=RM

By distance formula: p

(cosα−cosβ)2_{+ (sinα−sinβ)}2₌p

(cos(α−β)−1)2_{+ (sin(α−β)−0)}2

Square and expand the LHS:

cos2α−2 cosαcosβ+ cos2β+ sin2α−2 sinαsinβ+ sin2β

= 1 + 1−2(cosαcosβ+ sinαsinβ) = 2−2(cosαcosβ+ sinαsinβ)

Square and expand the RHS:

P Q=RM

By distance formula: p

(cosα−cosβ)2_{+ (sinα−sinβ)}2₌p

(cos(α−β)−1)2_{+ (sin(α−β)−0)}2

Square and expand the LHS:

cos2α−2 cosαcosβ+ cos2β+ sin2α−2 sinαsinβ+ sin2β

= 1 + 1−2(cosαcosβ+ sinαsinβ) = 2−2(cosαcosβ+ sinαsinβ)

Square and expand the RHS:

P Q=RM

By distance formula: p

(cosα−cosβ)2_{+ (sinα−sinβ)}2₌p

(cos(α−β)−1)2_{+ (sin(α−β)−0)}2

Square and expand the LHS:

cos2α−2 cosαcosβ+ cos2β+ sin2α−2 sinαsinβ+ sin2β

= 1

+ 1−2(cosαcosβ+ sinαsinβ) = 2−2(cosαcosβ+ sinαsinβ)

Square and expand the RHS:

P Q=RM

By distance formula: p

(cosα−cosβ)2_{+ (sinα−sinβ)}2₌p

(cos(α−β)−1)2_{+ (sin(α−β)−0)}2

Square and expand the LHS:

cos2α−2 cosαcosβ+ cos2β+ sin2α−2 sinαsinβ+ sin2β

= 1 + 1

−2(cosαcosβ+ sinαsinβ) = 2−2(cosαcosβ+ sinαsinβ)

Square and expand the RHS:

P Q=RM

By distance formula: p

(cosα−cosβ)2_{+ (sinα−sinβ)}2₌p

(cos(α−β)−1)2_{+ (sin(α−β)−0)}2

Square and expand the LHS:

cos2α−2 cosαcosβ+ cos2β+ sin2α−2 sinαsinβ+ sin2β

= 1 + 1−2(cosαcosβ+ sinαsinβ)

= 2−2(cosαcosβ+ sinαsinβ)

Square and expand the RHS:

P Q=RM

By distance formula: p

(cosα−cosβ)2_{+ (sinα−sinβ)}2₌p

(cos(α−β)−1)2_{+ (sin(α−β)−0)}2

Square and expand the LHS:

cos2α−2 cosαcosβ+ cos2β+ sin2α−2 sinαsinβ+ sin2β

= 1 + 1−2(cosαcosβ+ sinαsinβ) = 2−2(cosαcosβ+ sinαsinβ)

Square and expand the RHS:

P Q=RM

By distance formula: p

(cosα−cosβ)2_{+ (sinα−sinβ)}2₌p

(cos(α−β)−1)2_{+ (sin(α−β)−0)}2

Square and expand the LHS:

cos2α−2 cosαcosβ+ cos2β+ sin2α−2 sinαsinβ+ sin2β

= 1 + 1−2(cosαcosβ+ sinαsinβ) = 2−2(cosαcosβ+ sinαsinβ)

Square and expand the RHS:

P Q=RM

By distance formula: p

(cosα−cosβ)2_{+ (sinα−sinβ)}2₌p

(cos(α−β)−1)2_{+ (sin(α−β)−0)}2

Square and expand the LHS:

cos2α−2 cosαcosβ+ cos2β+ sin2α−2 sinαsinβ+ sin2β

= 1 + 1−2(cosαcosβ+ sinαsinβ) = 2−2(cosαcosβ+ sinαsinβ)

Square and expand the RHS:

cos2(α−β)−2 cos(α−β) + 1 + sin2(α−β)

P Q=RM

By distance formula: p

(cosα−cosβ)2_{+ (sinα−sinβ)}2₌p

(cos(α−β)−1)2_{+ (sin(α−β)−0)}2

Square and expand the LHS:

cos2α−2 cosαcosβ+ cos2β+ sin2α−2 sinαsinβ+ sin2β

= 1 + 1−2(cosαcosβ+ sinαsinβ) = 2−2(cosαcosβ+ sinαsinβ)

Square and expand the RHS:

cos2(α−β)−2 cos(α−β)+ 1 + sin2(α−β)

2−2(cosαcosβ+ sinαsinβ) = 2−2 cos(α−β)

2 cos(α−β) = 2(cosαcosβ+ sinαsinβ)

cos(α−β) = cosαcosβ+ sinαsinβ

cos(α+β) = cos[α−(−β)]

= cosαcos(−β) + sinαsin(−β) = cosαcosβ−sinαsinβ

2−2(cosαcosβ+ sinαsinβ)= 2−2 cos(α−β)

2 cos(α−β) = 2(cosαcosβ+ sinαsinβ)

cos(α−β) = cosαcosβ+ sinαsinβ

cos(α+β) = cos[α−(−β)]

= cosαcos(−β) + sinαsin(−β) = cosαcosβ−sinαsinβ

2−2(cosαcosβ+ sinαsinβ) = 2−2 cos(α−β)

2 cos(α−β) = 2(cosαcosβ+ sinαsinβ)

cos(α−β) = cosαcosβ+ sinαsinβ

cos(α+β) = cos[α−(−β)]

= cosαcos(−β) + sinαsin(−β) = cosαcosβ−sinαsinβ

2−2(cosαcosβ+ sinαsinβ) = 2−2 cos(α−β)

2 cos(α−β) = 2(cosαcosβ+ sinαsinβ)

cos(α−β) = cosαcosβ+ sinαsinβ

cos(α+β)

= cos[α−(−β)]

= cosαcos(−β) + sinαsin(−β) = cosαcosβ−sinαsinβ

2−2(cosαcosβ+ sinαsinβ) = 2−2 cos(α−β)

2 cos(α−β) = 2(cosαcosβ+ sinαsinβ)

cos(α−β) = cosαcosβ+ sinαsinβ

cos(α+β) = cos[α−(−β)]

= cosαcos(−β) + sinαsin(−β) = cosαcosβ−sinαsinβ

2−2(cosαcosβ+ sinαsinβ) = 2−2 cos(α−β)

2 cos(α−β) = 2(cosαcosβ+ sinαsinβ)

cos(α−β) = cosαcosβ+ sinαsinβ

cos(α+β) = cos[α−(−β)]

= cosαcos(−β) + sinαsin(−β)

= cosαcosβ−sinαsinβ

2−2(cosαcosβ+ sinαsinβ) = 2−2 cos(α−β)

2 cos(α−β) = 2(cosαcosβ+ sinαsinβ)

cos(α−β) = cosαcosβ+ sinαsinβ

cos(α+β) = cos[α−(−β)]

= cosαcos(−β)+ sinαsin(−β) = cosαcosβ

−sinαsinβ

2−2(cosαcosβ+ sinαsinβ) = 2−2 cos(α−β)

2 cos(α−β) = 2(cosαcosβ+ sinαsinβ)

cos(α−β) = cosαcosβ+ sinαsinβ

cos(α+β) = cos[α−(−β)]

= cosαcos(−β) + sinαsin(−β)

= cosαcosβ−sinαsinβ

2−2(cosαcosβ+ sinαsinβ) = 2−2 cos(α−β)

2 cos(α−β) = 2(cosαcosβ+ sinαsinβ)

cos(α−β) = cosαcosβ+ sinαsinβ

cos(α+β) = cos[α−(−β)]

= cosαcos(−β) + sinαsin(−β) = cosαcosβ−sinαsinβ

Example: Evaluatecosπ 12

. Solution:

cosπ 12

= cosπ 3 −

π

4

= cosπ 3cos

π

4 + sin

π 3sin π 4 = 1 2· √ 2 2 + √ 3 2 · √ 2 2 = √ 2 4 + √ 6 4 = √

Example: Evaluatecosπ 12

. Solution:

cosπ 12

= cosπ 3 −

π

4

= cosπ 3cos

π

4 + sin

π 3sin π 4 = 1 2· √ 2 2 + √ 3 2 · √ 2 2 = √ 2 4 + √ 6 4 = √

Example: Evaluatecosπ 12

. Solution:

cosπ 12

= cosπ 3 −

π

4

= cosπ 3cos

π

4 + sin

π 3sin π 4 = 1 2· √ 2 2 + √ 3 2 · √ 2 2 = √ 2 4 + √ 6 4 = √

Example: Evaluatecosπ 12

. Solution:

cosπ 12

= cosπ 3 −

π

4

= cosπ 3cos

π

4 + sin

π 3sin π 4 = 1 2· √ 2 2 + √ 3 2 · √ 2 2 = √ 2 4 + √ 6 4 = √

Example: Evaluatecosπ 12

. Solution:

cosπ 12

= cosπ 3 −

π

4

= cosπ 3cos

π

4 + sin

π 3sin π 4 = 1 2· √ 2 2 + √ 3 2 · √ 2 2 = √ 2 4 + √ 6 4 = √

Example: Evaluatecosπ 12

. Solution:

cosπ 12

= cosπ 3 −

π

4

= cosπ 3cos

π

4 + sin

π 3sin π 4 = 1 2· √ 2 2 + √ 3 2 · √ 2 2 = √ 2 4 + √ 6 4 = √

Example: Evaluatecosπ 12

. Solution:

cosπ 12

= cosπ 3 −

π

4

= cosπ 3cos

π

4 + sin

π 3sin π 4 = 1 2· √ 2 2 + √ 3 2 · √ 2 2 = √ 2 4 + √ 6 4 = √

Example: Evaluatecosπ 12

. Solution:

cosπ 12

= cosπ 3 −

π

4

= cosπ 3cos

π

4 + sin

π 3sin π 4 = 1 2· √ 2 2 + √ 3 2 · √ 2 2 = √ 2 4 + √ 6 4 = √

Example: Evaluatecosπ 12

. Solution:

cosπ 12

= cosπ 3 −

π

4

= cosπ 3cos

π

4 + sin

π 3sin π 4 = 1 2· √ 2 2 + √ 3 2 · √ 2 2 = √ 2 4 + √ 6 4 = √

cos π

2 −α

= cosπ

2cosα+ sin

π

2sinα = 0·cosα+ 1·sinα

cosπ 2 −α

= sinα

Replacing α by π

2 −α:

sin π

2 −α

= cos hπ

2 − π

2 −α i

= cos π

2 −

π

2 +α

sinπ 2 −α

cos π

2 −α

= cosπ

2cosα+ sin

π

2sinα

= 0·cosα+ 1·sinα

cosπ 2 −α

= sinα

Replacing α by π

2 −α:

sin π

2 −α

= cos hπ

2 − π

2 −α i

= cos π

2 −

π

2 +α

sinπ 2 −α

cos π

2 −α

= cosπ

2cosα+ sin

π

2sinα =0·cosα

+ 1·sinα

cosπ 2 −α

= sinα

Replacing α by π

2 −α:

sin π

2 −α

= cos hπ

2 − π

2 −α i

= cos π

2 −

π

2 +α

sinπ 2 −α

cos π

2 −α

= cosπ

2cosα+ sin

π 2sinα

= 0·cosα+1·sinα

cosπ 2 −α

= sinα

Replacing α by π

2 −α:

sin π

2 −α

= cos hπ

2 − π

2 −α i

= cos π

2 −

π

2 +α

sinπ 2 −α

cos π

2 −α

= cosπ

2cosα+ sin

π

2sinα = 0·cosα+ 1·sinα

cosπ 2 −α

= sinα

Replacing α by π

2 −α:

sin π

2 −α

= cos hπ

2 − π

2 −α i

= cos π

2 −

π

2 +α

sinπ 2 −α

cos π

2 −α

= cosπ

2cosα+ sin

π

2sinα = 0·cosα+ 1·sinα

cosπ 2 −α

= sinα

Replacing α by π

2 −α:

sin π

2 −α

= cos hπ

2 − π

2 −α i

= cos π

2 −

π

2 +α

sinπ 2 −α

cos π

2 −α

= cosπ

2cosα+ sin

π

2sinα = 0·cosα+ 1·sinα

cosπ 2 −α

= sinα

Replacing α by π 2 −α:

sin π

2 −α

=

cos hπ

2 − π

2 −α i

= cos π

2 −

π

2 +α

sinπ 2 −α

cos π

2 −α

= cosπ

2cosα+ sin

π

2sinα = 0·cosα+ 1·sinα

cosπ 2 −α

= sinα

Replacing α by π 2 −α:

sin π

2 −α

= cos hπ

2 − π

2 −α i

=

cos π

2 −

π

2 +α

sinπ 2 −α

cos π

2 −α

= cosπ

2cosα+ sin

π

2sinα = 0·cosα+ 1·sinα

cosπ 2 −α

= sinα

Replacing α by π

2 −α:

sin π

2 −α

= cos hπ

2 − π

2 −α

i = cos

π

2 − π 2 +α

sinπ 2 −α

cos π

2 −α

= cosπ

2cosα+ sin

π

2sinα = 0·cosα+ 1·sinα

cosπ 2 −α

= sinα

Replacing α by π

2 −α:

sin π

2 −α

= cos hπ

2 − π

2 −α i

= cos π

2 −

π

2 +α

sinπ 2 −α

Other Sum/Difference Identities

sin(α+β) = cosπ

2 −(α+β)

= coshπ 2 −α

−βi

= cos π

2 −α

cosβ+ sin π

2 −α

sinβ

= sinαcosβ+ cosαsinβ

sin(α+β) = sinαcosβ+ cosαsinβ

Similarly,

Other Sum/Difference Identities

sin(α+β) = cosπ

2 −(α+β)

= coshπ 2 −α

−βi

= cos π

2 −α

cosβ+ sin π

2 −α

sinβ

= sinαcosβ+ cosαsinβ

sin(α+β) = sinαcosβ+ cosαsinβ

Similarly,

Other Sum/Difference Identities

sin(α+β) = cosπ

2 −(α+β)

= coshπ 2 −α

−βi

= cos π

2 −α

cosβ+ sin π

2 −α

sinβ

= sinαcosβ+ cosαsinβ

sin(α+β) = sinαcosβ+ cosαsinβ

Similarly,

Other Sum/Difference Identities

sin(α+β) = cosπ

2 −(α+β)

= coshπ 2 −α

−βi

= cos π

2 −α

cosβ+ sin π

2 −α

sinβ

= sinαcosβ+

cosαsinβ

sin(α+β) = sinαcosβ+ cosαsinβ

Similarly,

Other Sum/Difference Identities

sin(α+β) = cosπ

2 −(α+β)

= coshπ 2 −α

−βi

= cos π

2 −α

cosβ+ sin π

2 −α

sinβ

= sinαcosβ+ cosαsinβ

sin(α+β) = sinαcosβ+ cosαsinβ

Similarly,

Other Sum/Difference Identities

sin(α+β) = cosπ

2 −(α+β)

= coshπ 2 −α

−βi

= cos π

2 −α

cosβ+ sin π

2 −α

sinβ

= sinαcosβ+ cosαsinβ

sin(α+β) = sinαcosβ+ cosαsinβ

Similarly,

Other Sum/Difference Identities

sin(α+β) = cosπ

2 −(α+β)

= coshπ 2 −α

−βi

= cos π

2 −α

cosβ+ sin π

2 −α

sinβ

= sinαcosβ+ cosαsinβ

sin(α+β) = sinαcosβ+ cosαsinβ

Similarly,

Example: Evaluatesin 195◦. Solution:

sin 195◦ = sin(135◦+ 60◦)

= sin 135◦cos 60◦+ cos 135◦sin 60◦

=

√

2 2

1

2

+

− √

2 2

√

3 2

=

√

Example: Evaluatesin 195◦. Solution:

sin 195◦ =

sin(135◦+ 60◦)

= sin 135◦cos 60◦+ cos 135◦sin 60◦

=

√

2 2

1

2

+

− √

2 2

√

3 2

=

√

Example: Evaluatesin 195◦. Solution:

sin 195◦ = sin(135◦+ 60◦)

= sin 135◦cos 60◦+ cos 135◦sin 60◦

=

√

2 2

1

2

+

− √

2 2

√

3 2

=

√

Example: Evaluatesin 195◦. Solution:

sin 195◦ = sin(135◦+ 60◦) = sin 135◦cos 60◦

+ cos 135◦sin 60◦

=

√

2 2

1

2

+

− √

2 2

√

3 2

=

√

Example: Evaluatesin 195◦. Solution:

sin 195◦ = sin(135◦+ 60◦)

= sin 135◦cos 60◦+ cos 135◦sin 60◦

=

√

2 2

1

2

+

− √

2 2

√

3 2

=

√

Example: Evaluatesin 195◦. Solution:

sin 195◦ = sin(135◦+ 60◦)

= sin 135◦cos 60◦+ cos 135◦sin 60◦

=

√

2 2

1

2

+

− √

2 2

√

3 2

=

√

Example: Evaluatesin 195◦. Solution:

sin 195◦ = sin(135◦+ 60◦)

= sin 135◦cos 60◦+ cos 135◦sin 60◦

=

√

2 2

1

2

+

− √

2 2

√

3 2

=

√

Example: Evaluatesin 195◦. Solution:

sin 195◦ = sin(135◦+ 60◦)

= sin 135◦cos 60◦+ cos 135◦sin 60◦

=

√

2 2

1

2

+

− √

2 2

√

3 2

=

√

Example: Evaluatesin 195◦. Solution:

sin 195◦ = sin(135◦+ 60◦)

= sin 135◦cos 60◦+ cos 135◦sin 60◦

=

√

2 2

1

2

+

− √

2 2

√

3 2

=

√

Example: Evaluatesin 195◦. Solution:

sin 195◦ = sin(135◦+ 60◦)

= sin 135◦cos 60◦+ cos 135◦sin 60◦

=

√

2 2

1

2

+

− √

2 2

√

3 2

=

√

tan(α+β)

= sin(α+β) cos(α+β)

= sinαcosβ+ cosαsinβ cosαcosβ− sinαsinβ ·

1 cosαcosβ

1 cosαcosβ

=

sinα

cosα+

sinβ

cosβ

1− sinαsinβ

cosαcosβ

tan(α+β) = tanα+ tanβ 1−tanαtanβ

Similarly,

tan(α+β) = sin(α+β) cos(α+β)

= sinαcosβ+ cosαsinβ cosαcosβ− sinαsinβ ·

1 cosαcosβ

1 cosαcosβ

=

sinα

cosα+

sinβ

cosβ

1− sinαsinβ

cosαcosβ

tan(α+β) = tanα+ tanβ 1−tanαtanβ

Similarly,

tan(α+β) = sin(α+β) cos(α+β)

= sinαcosβ+ cosαsinβ

cosαcosβ− sinαsinβ ·

1 cosαcosβ

1 cosαcosβ

=

sinα

cosα+

sinβ

cosβ

1− sinαsinβ

cosαcosβ

tan(α+β) = tanα+ tanβ 1−tanαtanβ

Similarly,

tan(α+β) = sin(α+β) cos(α+β)

= sinαcosβ+ cosαsinβ cosαcosβ− sinαsinβ

·

1 cosαcosβ

1 cosαcosβ

=

sinα

cosα+

sinβ

cosβ

1− sinαsinβ

cosαcosβ

tan(α+β) = tanα+ tanβ 1−tanαtanβ

Similarly,

tan(α+β) = sin(α+β) cos(α+β)

= sinαcosβ+ cosαsinβ cosαcosβ− sinαsinβ ·

1 cosαcosβ

1 cosαcosβ

=

sinα

cosα+

sinβ

cosβ

1− sinαsinβ

cosαcosβ

tan(α+β) = tanα+ tanβ 1−tanαtanβ

Similarly,

tan(α+β) = sin(α+β) cos(α+β)

= sinαcosβ+ cosαsinβ cosαcosβ− sinαsinβ ·

1 cosαcosβ

1 cosαcosβ

=

sinα

cosα

+ _cossinβ_β

1− sinαsinβ

cosαcosβ

tan(α+β) = tanα+ tanβ 1−tanαtanβ

Similarly,

tan(α+β) = sin(α+β) cos(α+β)

= sinαcosβ+ cosαsinβ cosαcosβ− sinαsinβ ·

1 cosαcosβ

1 cosαcosβ

=

sinα

cosα+

sinβ

cosβ

1− sinαsinβ

cosαcosβ

tan(α+β) = tanα+ tanβ 1−tanαtanβ

Similarly,

tan(α+β) = sin(α+β) cos(α+β)

= sinαcosβ+ cosαsinβ

cosαcosβ− sinαsinβ ·

1 cosαcosβ

1 cosαcosβ

=

sinα

cosα+

sinβ

cosβ

1

− sinαsinβ

cosαcosβ

tan(α+β) = tanα+ tanβ 1−tanαtanβ

Similarly,

tan(α+β) = sin(α+β) cos(α+β)

= sinαcosβ+ cosαsinβ cosαcosβ− sinαsinβ ·

1 cosαcosβ

1 cosαcosβ

=

sinα

cosα+

sinβ

cosβ

1− sinαsinβ

cosαcosβ

tan(α+β) = tanα+ tanβ 1−tanαtanβ

Similarly,

tan(α+β) = sin(α+β) cos(α+β)

= sinαcosβ+ cosαsinβ cosαcosβ− sinαsinβ ·

1 cosαcosβ

1 cosαcosβ

=

sinα

cosα+

sinβ

cosβ

1− sinαsinβ

cosαcosβ

tan(α+β) = tanα+ tanβ 1−tanαtanβ

Similarly,

tan(α+β) = sin(α+β) cos(α+β)

= sinαcosβ+ cosαsinβ cosαcosβ− sinαsinβ ·

1 cosαcosβ

1 cosαcosβ

=

sinα

cosα+

sinβ

cosβ

1− sinαsinβ

cosαcosβ

tan(α+β) = tanα+ tanβ 1−tanαtanβ

Similarly,

Example: Evaluate tan 320

◦_{−tan 95}◦

1 + tan 320◦_{tan 95}◦.

Solution:

tan 320◦ −tan 95◦

1 + tan 320◦_{tan 95}◦ = tan(320 ◦₋

Example: Evaluate tan 320

◦_{−tan 95}◦

1 + tan 320◦_{tan 95}◦.

Solution:

tan 320◦−tan 95◦ 1 + tan 320◦_{tan 95}◦ =

Example: Evaluate tan 320

◦_{−tan 95}◦

1 + tan 320◦_{tan 95}◦.

Solution:

tan 320◦−tan 95◦

1 + tan 320◦_{tan 95}◦ = tan(320 ◦₋

95◦)

Example: Evaluate tan 320

◦_{−tan 95}◦

1 + tan 320◦_{tan 95}◦.

Solution:

tan 320◦−tan 95◦

1 + tan 320◦_{tan 95}◦ = tan(320 ◦₋

95◦) = tan(225◦)

Example: Evaluate tan 320

◦_{−tan 95}◦

1 + tan 320◦_{tan 95}◦.

Solution:

tan 320◦−tan 95◦

1 + tan 320◦_{tan 95}◦ = tan(320 ◦₋

Co-function Identities

In addition to cos π

2 −α

= sinα and sin π

2 −α

= cosα,

tanπ 2 −α

=

sinπ 2 −α

cosπ 2 −α

=

cosα

sinα = cotα

cotπ 2 −α

= tanα

secπ 2 −α

= cscα

cscπ 2 −α

= secα

We say that sinand cos, tan and cot, and secand csc are

Co-function Identities

In addition to cos π

2 −α

= sinα and sin π

2 −α

= cosα,

tanπ 2 −α

=

sinπ 2 −α

cosπ 2 −α

=

cosα

sinα = cotα

cotπ 2 −α

= tanα

secπ 2 −α

= cscα

cscπ 2 −α

= secα

We say that sinand cos, tan and cot, and secand csc are

Co-function Identities

In addition to cos π

2 −α

= sinα and sin π

2 −α

= cosα,

tanπ 2 −α

=

sinπ 2 −α

cosπ 2 −α

=

cosα

sinα = cotα

cotπ 2 −α

= tanα

secπ 2 −α

= cscα

cscπ 2 −α

= secα

We say that sinand cos, tan and cot, and secand csc are

Co-function Identities

In addition to cos π

2 −α

= sinα and sin π

2 −α

= cosα,

tanπ 2 −α

=

sinπ 2 −α

cosπ 2 −α

=

cosα

sinα

= cotα

cotπ 2 −α

= tanα

secπ 2 −α

= cscα

cscπ 2 −α

= secα

We say that sinand cos, tan and cot, and secand csc are

Co-function Identities

In addition to cos π

2 −α

= sinα and sin π

2 −α

= cosα,

tanπ 2 −α

=

sinπ 2 −α

cosπ 2 −α

=

cosα

sinα = cotα

cotπ 2 −α

= tanα

secπ 2 −α

= cscα

cscπ 2 −α

= secα

We say that sinand cos, tan and cot, and secand csc are

Co-function Identities

In addition to cos π

2 −α

= sinα and sin π

2 −α

= cosα,

tanπ 2 −α

=

sinπ 2 −α

cosπ 2 −α

=

cosα

sinα = cotα

cot π

2 −α

= tanα

secπ 2 −α

= cscα

cscπ 2 −α

= secα

We say that sinand cos, tan and cot, and secand csc are

Co-function Identities

In addition to cos π

2 −α

= sinα and sin π

2 −α

= cosα,

tanπ 2 −α

=

sinπ 2 −α

cosπ 2 −α

=

cosα

sinα = cotα

cot π

2 −α

= tanα

secπ 2 −α

= cscα

cscπ 2 −α

= secα

We say that sinand cos, tan and cot, and secand csc are

Co-Function Identities

sinπ 2 −α

= cosα

cosπ 2 −α

= sinα

tanπ 2 −α

= cotα

cotπ 2 −α

= tanα

secπ 2 −α

= cscα

cscπ 2 −α

= secα

Note:

Co-Function Identities

sinπ 2 −α

= cosα

cosπ 2 −α

= sinα

tanπ 2 −α

= cotα

cotπ 2 −α

= tanα

secπ 2 −α

= cscα

cscπ 2 −α

= secα

Note:

Example: Giventanα= 3₄ such that α does not lie in Quadrant I, and sinβ = ₁₃5 such that cotβ <0. Find:

1 csc(π

2 −β) 2 sin(α−β) 3 cot(α+β)

Solution: α lies in QIII

x y _r

Takey=−3, x=−4. r2 = (−4)2+ (−3)2

r=√16 + 9

= 5

sinα

=−3 5

, cosα =−4 5

From the Pythagorean identity,

cos2_{β = 1−(}5 13)

2

cos2β = 1−(₁₆₉25) cos2_{β =} 144

169

cosβ =−12

13

tanβ =− 5

Example: Giventanα= 3₄ such that α does not lie in Quadrant I, and sinβ = ₁₃5 such that cotβ <0. Find:

1 csc(π

2 −β) 2 sin(α−β) 3 cot(α+β)

Solution: α lies in QIII

x y _r

Takey=−3,x=−4. r2 = (−4)2+ (−3)2

r=√16 + 9

= 5

sinα

=−3 5

, cosα=−4 5

From the Pythagorean identity,

cos2_{β = 1−(}5 13)

2

cos2β = 1−(₁₆₉25) cos2_{β =} 144

169

cosβ =−12

13

tanβ =− 5

Example: Giventanα= 3₄ such that α does not lie in Quadrant I, and sinβ = ₁₃5 such that cotβ <0. Find:

1 csc(π

2 −β) 2 sin(α−β) 3 cot(α+β)

Solution: α lies in QIII

x

y _r

Takey=−3,x=−4. r2 = (−4)2+ (−3)2

r=√16 + 9

= 5

sinα

=−3 5

, cosα=−4 5

From the Pythagorean identity,

cos2_{β = 1−(}5 13)

2

cos2β = 1−(₁₆₉25) cos2_{β =} 144

169

cosβ =−12

13

tanβ =− 5

Example: Giventanα= 3₄ such that α does not lie in Quadrant I, and sinβ = ₁₃5 such that cotβ <0. Find:

1 csc(π

2 −β) 2 sin(α−β) 3 cot(α+β)

Solution: α lies in QIII

x

y _r

Takey=−3,x=−4.

r2 = (−4)2+ (−3)2

r=√16 + 9

= 5

sinα

=−3 5

, cosα=−4 5

From the Pythagorean identity,

cos2_{β = 1−(}5 13)

2

cos2β = 1−(₁₆₉25) cos2_{β =} 144

169

cosβ =−12

13

tanβ =− 5

Example: Giventanα= 3₄ such that α does not lie in Quadrant I, and sinβ = ₁₃5 such that cotβ <0. Find:

1 csc(π

2 −β) 2 sin(α−β) 3 cot(α+β)

Solution: α lies in QIII

x

y _r

Takey=−3,x=−4. r2 = (−4)2+ (−3)2

r=√16 + 9

= 5

sinα

=−3 5

, cosα=−4 5

From the Pythagorean identity,

cos2_{β = 1−(}5 13)

2

cos2β = 1−(₁₆₉25) cos2_{β =} 144

169

cosβ =−12

13

tanβ =− 5

Example: Giventanα= 3₄ such that α does not lie in Quadrant I, and sinβ = ₁₃5 such that cotβ <0. Find:

1 csc(π

2 −β) 2 sin(α−β) 3 cot(α+β)

Solution: α lies in QIII

x

y _r

Takey=−3,x=−4. r2 = (−4)2+ (−3)2

r=√16 + 9

= 5 sinα

=−3 5

, cosα=−4 5

From the Pythagorean identity,

cos2_{β = 1−(}5 13)

2

cos2β = 1−(₁₆₉25) cos2_{β =} 144

169

cosβ =−12

13

tanβ =− 5

Example: Giventanα= 3₄ such that α does not lie in Quadrant I, and sinβ = ₁₃5 such that cotβ <0. Find:

1 csc(π

2 −β) 2 sin(α−β) 3 cot(α+β)

Solution: α lies in QIII

x

y _r

Takey=−3,x=−4. r2 = (−4)2+ (−3)2

r=√16 + 9 = 5

sinα

=−3 5

, cosα=−4 5

From the Pythagorean identity,

cos2_{β = 1−(}5 13)

2

cos2β = 1−(₁₆₉25) cos2_{β =} 144

169

cosβ =−12

13

tanβ =− 5

Example: Giventanα= 3₄ such that α does not lie in Quadrant I, and sinβ = ₁₃5 such that cotβ <0. Find:

1 csc(π

2 −β) 2 sin(α−β) 3 cot(α+β)

Solution: α lies in QIII

x

y _r

Takey=−3,x=−4. r2 = (−4)2+ (−3)2

r=√16 + 9 = 5 sinα

=−3 5

,

cosα=−4 5

From the Pythagorean identity,

cos2_{β = 1−(}5 13)

2

cos2β = 1−(₁₆₉25) cos2_{β =} 144

169

cosβ =−12

13

tanβ =− 5

Example: Giventanα= 3₄ such that α does not lie in Quadrant I, and sinβ = ₁₃5 such that cotβ <0. Find:

1 csc(π

2 −β) 2 sin(α−β) 3 cot(α+β)

Solution: α lies in QIII

x

y _r

Takey=−3,x=−4. r2 = (−4)2+ (−3)2

r=√16 + 9 = 5 sinα=−3

5,

cosα=−4 5

From the Pythagorean identity,

cos2_{β = 1−(}5 13)

2

cos2β = 1−(₁₆₉25) cos2_{β =} 144

169

cosβ =−12

13

tanβ =− 5

Example: Giventanα= 3₄ such that α does not lie in Quadrant I, and sinβ = ₁₃5 such that cotβ <0. Find:

1 csc(π

2 −β) 2 sin(α−β) 3 cot(α+β)

Solution: α lies in QIII

x

y _r

Takey=−3,x=−4. r2 = (−4)2+ (−3)2

r=√16 + 9 = 5 sinα=−3

5, cosα=− 4 5

From the Pythagorean identity,

cos2_{β = 1−(}5 13)

2

cos2β = 1−(₁₆₉25) cos2_{β =} 144

169

cosβ =−12

13

tanβ =− 5

Example: Giventanα= 3₄ such that α does not lie in Quadrant I, and sinβ = ₁₃5 such that cotβ <0. Find:

1 csc(π

2 −β) 2 sin(α−β) 3 cot(α+β)

Solution: α lies in QIII

x

y _r

Takey=−3,x=−4. r2 = (−4)2+ (−3)2

r=√16 + 9 = 5 sinα=−3

5, cosα=− 4 5

From the Pythagorean identity,

cos2_{β = 1−(}5 13)

2

cos2β = 1−(₁₆₉25) cos2_{β =} 144

169

cosβ =−12

13

tanβ =− 5

Example: Giventanα= 3₄ such that α does not lie in Quadrant I, and sinβ = ₁₃5 such that cotβ <0. Find:

1 csc(π

2 −β) 2 sin(α−β) 3 cot(α+β)

Solution: α lies in QIII

x

y _r

Takey=−3,x=−4. r2 = (−4)2+ (−3)2

r=√16 + 9 = 5 sinα=−3

5, cosα=− 4 5

From the Pythagorean identity,

cos2_{β = 1−(}5 13)

2

cos2β = 1−(₁₆₉25)

cos2_{β =} 144 169

cosβ =−12

13

tanβ =− 5

Example: Giventanα= 3₄ such that α does not lie in Quadrant I, and sinβ = ₁₃5 such that cotβ <0. Find:

1 csc(π

2 −β) 2 sin(α−β) 3 cot(α+β)

Solution: α lies in QIII

x

y _r

Takey=−3,x=−4. r2 = (−4)2+ (−3)2

r=√16 + 9 = 5 sinα=−3

5, cosα=− 4 5

From the Pythagorean identity,

cos2_{β = 1−(}5 13)

2

cos2β = 1−(₁₆₉25) cos2β = 144₁₆₉

cosβ =−12

13

tanβ =− 5

Example: Giventanα= 3₄ such that α does not lie in Quadrant I, and sinβ = ₁₃5 such that cotβ <0. Find:

1 csc(π

2 −β) 2 sin(α−β) 3 cot(α+β)

Solution: α lies in QIII

x

y _r

Takey=−3,x=−4. r2 = (−4)2+ (−3)2

r=√16 + 9 = 5 sinα=−3

5, cosα=− 4 5

From the Pythagorean identity,

cos2_{β = 1−(}5 13)

2

cos2β = 1−(₁₆₉25) cos2β = 144₁₆₉

cosβ=−12

13

tanβ =− 5

Example: Giventanα= 3₄ such that α does not lie in Quadrant I, and sinβ = ₁₃5 such that cotβ <0. Find:

1 csc(π

2 −β) 2 sin(α−β) 3 cot(α+β)

Solution: α lies in QIII

x

y _r

Takey=−3,x=−4. r2 = (−4)2+ (−3)2

r=√16 + 9 = 5 sinα=−3

5, cosα=− 4 5

From the Pythagorean identity,

cos2_{β = 1−(}5 13)

2

cos2β = 1−(₁₆₉25) cos2β = 144₁₆₉

cosβ=−12

13

tanβ =− 5

Example: Giventanα= 3₄ such that α does not lie in Quadrant I, and sinβ = ₁₃5 such that cotβ <0. Find:

1 csc(π

2 −β) 2 sin(α−β) 3 cot(α+β)

Solution (cont):

csc(π₂ −β)

= secβ =−13 12

sin(α−β) = sinαcosβ−cosαsinβ

= −3 5 − 12 13 − −4 5 5 13 = 36 65 + 20 65 = 56 65

cot(α+β) = 1

tan(α+β) =

1−tanαtanβ

tanα+ tanβ

= 1−(

3 4)(−

5 12) 3

4 + (− 5 12)

· 48

48 =

48 + 15 36−20 =

Example: Giventanα= 3₄ such that α does not lie in Quadrant I, and sinβ = ₁₃5 such that cotβ <0. Find:

1 csc(π

2 −β) 2 sin(α−β) 3 cot(α+β)

Solution (cont):

csc(π₂ −β) = secβ

=−13 12

sin(α−β) = sinαcosβ−cosαsinβ

= −3 5 − 12 13 − −4 5 5 13 = 36 65 + 20 65 = 56 65

cot(α+β) = 1

tan(α+β) =

1−tanαtanβ

tanα+ tanβ

= 1−(

3 4)(−

5 12) 3

4 + (− 5 12)

· 48

48 =

48 + 15 36−20 =

Example: Giventanα= 3₄ such that α does not lie in Quadrant I, and sinβ = ₁₃5 such that cotβ <0. Find:

1 csc(π

2 −β) 2 sin(α−β) 3 cot(α+β)

Solution (cont):

csc(π₂ −β)= secβ =−13 12

sin(α−β) = sinαcosβ−cosαsinβ

= −3 5 − 12 13 − −4 5 5 13 = 36 65 + 20 65 = 56 65

cot(α+β) = 1

tan(α+β) =

1−tanαtanβ

tanα+ tanβ

= 1−(

3 4)(−

5 12) 3

4 + (− 5 12)

· 48

48 =

48 + 15 36−20 =

Example: Giventanα= 3₄ such that α does not lie in Quadrant I, and sinβ = ₁₃5 such that cotβ <0. Find:

1 csc(π

2 −β) 2 sin(α−β) 3 cot(α+β)

Solution (cont):

csc(π₂ −β) = secβ =−13 12

sin(α−β) = sinαcosβ−cosαsinβ

= −3 5 − 12 13 − −4 5 5 13 = 36 65 + 20 65 = 56 65

cot(α+β) = 1

tan(α+β) =

1−tanαtanβ

tanα+ tanβ

= 1−(

3 4)(−

5 12) 3

4 + (− 5 12)

· 48

48 =

48 + 15 36−20 =

Example: Giventanα= 3₄ such that α does not lie in Quadrant I, and sinβ = ₁₃5 such that cotβ <0. Find:

1 csc(π

2 −β) 2 sin(α−β) 3 cot(α+β)

Solution (cont):

csc(π₂ −β) = secβ =−13 12

sin(α−β) = sinαcosβ−cosαsinβ

= −3 5 − 12 13 − −4 5 5 13 = 36 65 + 20 65 = 56 65

cot(α+β) = 1

tan(α+β) =

1−tanαtanβ

tanα+ tanβ

= 1−(

3 4)(−

5 12) 3

4 + (− 5 12)

· 48

48 =

48 + 15 36−20 =

Example: Giventanα= 3₄ such that α does not lie in Quadrant I, and sinβ = ₁₃5 such that cotβ <0. Find:

1 csc(π

2 −β) 2 sin(α−β) 3 cot(α+β)

Solution (cont):

csc(π₂ −β) = secβ =−13 12

sin(α−β) = sinαcosβ−cosαsinβ

= −3 5 − 12 13 − −4 5 5 13 = 36 65 + 20 65 = 56 65

cot(α+β) = 1

tan(α+β) =

1−tanαtanβ

tanα+ tanβ

= 1−(

3 4)(−

5 12) 3

4 + (− 5 12)

· 48

48 =

48 + 15 36−20 =

Example: Giventanα= 3₄ such that α does not lie in Quadrant I, and sinβ = ₁₃5 such that cotβ <0. Find:

1 csc(π

2 −β) 2 sin(α−β) 3 cot(α+β)

Solution (cont):

csc(π₂ −β) = secβ =−13 12

sin(α−β)= sinαcosβ−cosαsinβ

= −3 5 − 12 13 − −4 5 5 13 = 36 65 + 20 65 = 56 65

cot(α+β) = 1

tan(α+β) =

1−tanαtanβ

tanα+ tanβ

= 1−(

3 4)(−

5 12) 3

4 + (− 5 12)

· 48

48 =

48 + 15 36−20 =

Example: Giventanα= 3₄ such that α does not lie in Quadrant I, and sinβ = ₁₃5 such that cotβ <0. Find:

1 csc(π

2 −β) 2 sin(α−β) 3 cot(α+β)

Solution (cont):

csc(π₂ −β) = secβ =−13 12

sin(α−β) = sinαcosβ−cosαsinβ

= −3 5 − 12 13 − −4 5 5 13 = 36 65 + 20 65 = 56 65

cot(α+β) = 1

tan(α+β)

= 1−tanαtanβ tanα+ tanβ

= 1−(

3 4)(−

5 12) 3

4 + (− 5 12)

· 48

48 =

48 + 15 36−20 =

Example: Giventanα= 3₄ such that α does not lie in Quadrant I, and sinβ = ₁₃5 such that cotβ <0. Find:

1 csc(π

2 −β) 2 sin(α−β) 3 cot(α+β)

Solution (cont):

csc(π₂ −β) = secβ =−13 12

sin(α−β) = sinαcosβ−cosαsinβ

= −3 5 − 12 13 − −4 5 5 13 = 36 65 + 20 65 = 56 65

cot(α+β) = 1

tan(α+β) =

1−tanαtanβ

tanα+ tanβ

= 1−(

3 4)(−

5 12) 3

4 + (− 5 12)

· 48

48 =

48 + 15 36−20 =

Example: Giventanα= 3₄ such that α does not lie in Quadrant I, and sinβ = ₁₃5 such that cotβ <0. Find:

1 csc(π

2 −β) 2 sin(α−β) 3 cot(α+β)

Solution (cont):

csc(π₂ −β) = secβ =−13 12

sin(α−β) = sinαcosβ−cosαsinβ

= −3 5 − 12 13 − −4 5 5 13 = 36 65 + 20 65 = 56 65

cot(α+β) = 1

tan(α+β) =

1−tanαtanβ

tanα+ tanβ

= 1−(

3 4)(−

5 12) 3

4 + (− 5 12)

· 48

48 =

48 + 15 36−20 =

Example: Giventanα= 3₄ such that α does not lie in Quadrant I, and sinβ = ₁₃5 such that cotβ <0. Find:

1 csc(π

2 −β) 2 sin(α−β) 3 cot(α+β)

Solution (cont):

csc(π₂ −β) = secβ =−13 12

sin(α−β) = sinαcosβ−cosαsinβ

= −3 5 − 12 13 − −4 5 5 13 = 36 65 + 20 65 = 56 65

cot(α+β) = 1

tan(α+β) =

1−tanαtanβ

tanα+ tanβ

= 1−(

3 4)(−

5 12) 3

4 + (− 5 12)

· 48

48

= 48 + 15 36−20 =

Example: Giventanα= 3₄ such that α does not lie in Quadrant I, and sinβ = ₁₃5 such that cotβ <0. Find:

1 csc(π

2 −β) 2 sin(α−β) 3 cot(α+β)

Solution (cont):

csc(π₂ −β) = secβ =−13 12

sin(α−β) = sinαcosβ−cosαsinβ

= −3 5 − 12 13 − −4 5 5 13 = 36 65 + 20 65 = 56 65

cot(α+β) = 1

tan(α+β) =

1−tanαtanβ

tanα+ tanβ

= 1−(

3 4)(−

5 12) 3

4 + (− 5 12)

· 48

48 =

48 + 15 36−20

Example: Giventanα= 3₄ such that α does not lie in Quadrant I, and sinβ = ₁₃5 such that cotβ <0. Find:

1 csc(π

2 −β) 2 sin(α−β) 3 cot(α+β)

Solution (cont):

csc(π₂ −β) = secβ =−13 12

sin(α−β) = sinαcosβ−cosαsinβ

= −3 5 − 12 13 − −4 5 5 13 = 36 65 + 20 65 = 56 65

cot(α+β)= 1

tan(α+β) =

1−tanαtanβ

tanα+ tanβ

= 1−(

3 4)(−

5 12) 3

4 + (− 5 12)

· 48

48 =

48 + 15 36−20 =

Example: Giventanα= 3₄ such that α does not lie in Quadrant I, and sinβ = ₁₃5 such that cotβ <0. Find:

1 csc(π

2 −β) 2 sin(α−β) 3 cot(α+β)

Solution (cont):

csc(π₂ −β)= secβ =−13 12

sin(α−β)= sinαcosβ−cosαsinβ

= −3 5 − 12 13 − −4 5 5 13 = 36 65 + 20 65 = 56 65

cot(α+β)= 1

tan(α+β) =

1−tanαtanβ

tanα+ tanβ

= 1−(

3 4)(−

5 12) 3

4 + (− 5 12)

· 48

48 =

48 + 15 36−20 =

Double Measure Identities

sin 2θ =

sin(θ+θ) = sinθcosθ+ cosθsinθ= 2 sinθcosθ

cos 2θ = cos(θ+θ) = cosθcosθ−sinθsinθ = cos2_θ−sin2_θ

cos 2θ = (1−sin2θ)−sin2θ = 1−2 sin2θ

cos 2θ = cos2θ−(1−cos2θ) = 2 cos2θ−1

tan 2θ= tan(θ+θ) = tanθ+ tanθ 1−tanθtanθ =

2 tanθ

Double Measure Identities

sin 2θ = sin(θ+θ) =

sinθcosθ+ cosθsinθ= 2 sinθcosθ

cos 2θ = cos(θ+θ) = cosθcosθ−sinθsinθ = cos2_θ−sin2_θ

cos 2θ = (1−sin2θ)−sin2θ = 1−2 sin2θ

cos 2θ = cos2θ−(1−cos2θ) = 2 cos2θ−1

tan 2θ= tan(θ+θ) = tanθ+ tanθ 1−tanθtanθ =

2 tanθ

Double Measure Identities

sin 2θ = sin(θ+θ) = sinθcosθ+ cosθsinθ= 2 sinθcosθ

cos 2θ = cos(θ+θ) = cosθcosθ−sinθsinθ = cos2_θ−sin2_θ

cos 2θ = (1−sin2θ)−sin2θ = 1−2 sin2θ

cos 2θ = cos2θ−(1−cos2θ) = 2 cos2θ−1

tan 2θ= tan(θ+θ) = tanθ+ tanθ 1−tanθtanθ =

2 tanθ

Double Measure Identities

sin 2θ = sin(θ+θ) = sinθcosθ+ cosθsinθ= 2 sinθcosθ

cos 2θ=

cos(θ+θ) = cosθcosθ−sinθsinθ = cos2_θ−sin2_θ

cos 2θ = (1−sin2θ)−sin2θ = 1−2 sin2θ

cos 2θ = cos2θ−(1−cos2θ) = 2 cos2θ−1

tan 2θ= tan(θ+θ) = tanθ+ tanθ 1−tanθtanθ =

2 tanθ

Double Measure Identities

sin 2θ = sin(θ+θ) = sinθcosθ+ cosθsinθ= 2 sinθcosθ

cos 2θ= cos(θ+θ) =

cosθcosθ−sinθsinθ = cos2_θ−sin2_θ

cos 2θ = (1−sin2θ)−sin2θ = 1−2 sin2θ

cos 2θ = cos2θ−(1−cos2θ) = 2 cos2θ−1

tan 2θ= tan(θ+θ) = tanθ+ tanθ 1−tanθtanθ =

2 tanθ

Double Measure Identities

sin 2θ = sin(θ+θ) = sinθcosθ+ cosθsinθ= 2 sinθcosθ

cos 2θ= cos(θ+θ) = cosθcosθ−sinθsinθ = cos2_θ−sin2_θ

cos 2θ = (1−sin2θ)−sin2θ = 1−2 sin2θ

cos 2θ = cos2θ−(1−cos2θ) = 2 cos2θ−1

tan 2θ= tan(θ+θ) = tanθ+ tanθ 1−tanθtanθ =

2 tanθ

Double Measure Identities

sin 2θ = sin(θ+θ) = sinθcosθ+ cosθsinθ= 2 sinθcosθ

cos 2θ= cos(θ+θ) = cosθcosθ−sinθsinθ = cos2_θ−sin2_θ

cos 2θ=

(1−sin2θ)−sin2θ = 1−2 sin2θ

cos 2θ = cos2θ−(1−cos2θ) = 2 cos2θ−1

tan 2θ= tan(θ+θ) = tanθ+ tanθ 1−tanθtanθ =

2 tanθ

Double Measure Identities

sin 2θ = sin(θ+θ) = sinθcosθ+ cosθsinθ= 2 sinθcosθ

cos 2θ= cos(θ+θ) = cosθcosθ−sinθsinθ = cos2_θ−sin2_θ

cos 2θ= (1−sin2θ)−sin2θ

= 1−2 sin2θ

cos 2θ = cos2θ−(1−cos2θ) = 2 cos2θ−1

tan 2θ= tan(θ+θ) = tanθ+ tanθ 1−tanθtanθ =

2 tanθ

Double Measure Identities

sin 2θ = sin(θ+θ) = sinθcosθ+ cosθsinθ= 2 sinθcosθ

cos 2θ= cos(θ+θ) = cosθcosθ−sinθsinθ = cos2_θ−sin2_θ

cos 2θ= (1−sin2θ)−sin2θ = 1−2 sin2θ

cos 2θ = cos2θ−(1−cos2θ) = 2 cos2θ−1

tan 2θ= tan(θ+θ) = tanθ+ tanθ 1−tanθtanθ =

2 tanθ

Double Measure Identities

sin 2θ = sin(θ+θ) = sinθcosθ+ cosθsinθ= 2 sinθcosθ

cos 2θ= cos(θ+θ) = cosθcosθ−sinθsinθ = cos2_θ−sin2_θ

cos 2θ= (1−sin2θ)−sin2θ = 1−2 sin2θ

cos 2θ=

cos2θ−(1−cos2θ) = 2 cos2θ−1

tan 2θ= tan(θ+θ) = tanθ+ tanθ 1−tanθtanθ =

2 tanθ

Double Measure Identities

sin 2θ = sin(θ+θ) = sinθcosθ+ cosθsinθ= 2 sinθcosθ

cos 2θ= cos(θ+θ) = cosθcosθ−sinθsinθ = cos2_θ−sin2_θ

cos 2θ= (1−sin2θ)−sin2θ = 1−2 sin2θ

cos 2θ= cos2θ−(1−cos2θ)

= 2 cos2θ−1

tan 2θ= tan(θ+θ) = tanθ+ tanθ 1−tanθtanθ =

2 tanθ

Double Measure Identities

sin 2θ = sin(θ+θ) = sinθcosθ+ cosθsinθ= 2 sinθcosθ

cos 2θ= cos(θ+θ) = cosθcosθ−sinθsinθ = cos2_θ−sin2_θ

cos 2θ= (1−sin2θ)−sin2θ = 1−2 sin2θ

cos 2θ= cos2θ−(1−cos2θ) = 2 cos2θ−1

tan 2θ= tan(θ+θ) = tanθ+ tanθ 1−tanθtanθ =

2 tanθ

Double Measure Identities

sin 2θ = sin(θ+θ) = sinθcosθ+ cosθsinθ= 2 sinθcosθ

cos 2θ= cos(θ+θ) = cosθcosθ−sinθsinθ = cos2_θ−sin2_θ

cos 2θ= (1−sin2θ)−sin2θ = 1−2 sin2θ

cos 2θ= cos2θ−(1−cos2θ) = 2 cos2θ−1

tan 2θ=

tan(θ+θ) = tanθ+ tanθ 1−tanθtanθ =

2 tanθ

Double Measure Identities

sin 2θ = sin(θ+θ) = sinθcosθ+ cosθsinθ= 2 sinθcosθ

cos 2θ= cos(θ+θ) = cosθcosθ−sinθsinθ = cos2_θ−sin2_θ

cos 2θ= (1−sin2θ)−sin2θ = 1−2 sin2θ

cos 2θ= cos2θ−(1−cos2θ) = 2 cos2θ−1

tan 2θ= tan(θ+θ) =

tanθ+ tanθ

1−tanθtanθ =

2 tanθ

Double Measure Identities

sin 2θ = sin(θ+θ) = sinθcosθ+ cosθsinθ= 2 sinθcosθ

cos 2θ= cos(θ+θ) = cosθcosθ−sinθsinθ = cos2_θ−sin2_θ

cos 2θ= (1−sin2θ)−sin2θ = 1−2 sin2θ

cos 2θ= cos2θ−(1−cos2θ) = 2 cos2θ−1

tan 2θ= tan(θ+θ) = tanθ+ tanθ 1−tanθtanθ =

2 tanθ

Double Measure Identities

sin 2θ = sin(θ+θ) = sinθcosθ+ cosθsinθ= 2 sinθcosθ

cos 2θ= cos(θ+θ) = cosθcosθ−sinθsinθ = cos2_θ−sin2_θ

cos 2θ= (1−sin2θ)−sin2θ = 1−2 sin2θ

cos 2θ= cos2θ−(1−cos2θ) = 2 cos2θ−1

tan 2θ= tan(θ+θ) = tanθ+ tanθ 1−tanθtanθ =

2 tanθ

sin 2θ= 2 sinθcosθ

cos 2θ= cos2θ−sin2θ

= 1−2 sin2θ

= 2 cos2θ−1

Example: Ifcosθ = 2 5 and

5π

4 < θ < 7π

4 , determine the value of

cos 2θ.

Solution:

cos 2θ = 2 cos2θ−1 = 2 2

5 2

−1 = 8

Example: Ifcosθ = 2 5 and

5π

4 < θ < 7π

4 , determine the value of

cos 2θ.

Solution:

cos 2θ =

2 cos2θ−1 = 2 2

5 2

−1 = 8

Example: Ifcosθ = 2 5 and

5π

4 < θ < 7π

4 , determine the value of

cos 2θ.

Solution:

cos 2θ = 2 cos2θ−1 =

2 2

5 2

−1 = 8

Example: Ifcosθ = 2 5 and

5π

4 < θ < 7π

4 , determine the value of

cos 2θ.

Solution:

cos 2θ = 2 cos2θ−1 = 2 2

5 2

−1 =

8

Example: Ifcosθ = 2 5 and

5π

4 < θ < 7π

4 , determine the value of

cos 2θ.

Solution:

cos 2θ = 2 cos2θ−1 = 2 2

5 2

−1 = 8

25−1 =

− 17

Example: Ifcosθ = 2 5 and

5π

4 < θ < 7π

4 , determine the value of

cos 2θ.

Solution:

cos 2θ = 2 cos2θ−1 = 2 2

5 2

−1 = 8

Example: Ifcotθ =− 7

24 and θ ∈

−π

2,0

, find sin 2θ. Solution:

tanθ= 1

cotθ =−

24 7

Sincecsc2_{θ= cot}2_{θ+ 1}

cscθ =−

q

−7 24

2

+ 1 =−25

24 and sinθ =− 24 25

Sincesec2θ= 1 + tan2θ

secθ = q

−24 7

2

+ 1 = 25

7 and cosθ = 7 25

So, sin 2θ =2 cosθsinθ = 2 −24 25 7 25

Example: Ifcotθ =− 7

24 and θ ∈

−π

2,0

, find sin 2θ. Solution:

tanθ = 1

cotθ =−

24 7

Sincecsc2_{θ= cot}2_{θ+ 1}

cscθ =−

q

−7 24

2

+ 1 =−25

24 and sinθ =− 24 25

Sincesec2θ= 1 + tan2θ

secθ = q

−24 7

2

+ 1 = 25

7 and cosθ = 7 25

So, sin 2θ =2 cosθsinθ = 2 −24 25 7 25

Example: Ifcotθ =− 7

24 and θ ∈

−π

2,0

, find sin 2θ. Solution:

tanθ = 1

cotθ =−

24 7

Sincecsc2_{θ= cot}2_{θ+ 1}

cscθ =−

q

−7 24

2

+ 1 =−25

24 and sinθ =− 24 25

Sincesec2θ= 1 + tan2θ

secθ = q

−24 7

2

+ 1 = 25

7 and cosθ = 7 25

So, sin 2θ =2 cosθsinθ = 2 −24 25 7 25

Example: Ifcotθ =− 7

24 and θ ∈

−π 2,0

, find sin 2θ. Solution:

tanθ = 1

cotθ =−

24 7

Sincecsc2_{θ= cot}2_{θ+ 1}

cscθ =−

q

−7 24

2 + 1

=−25

24 and sinθ =− 24 25

Sincesec2θ= 1 + tan2θ

secθ = q

−24 7

2

+ 1 = 25

7 and cosθ = 7 25

So, sin 2θ =2 cosθsinθ = 2 −24 25 7 25

Example: Ifcotθ =− 7

24 and θ ∈

−π

2,0

, find sin 2θ. Solution:

tanθ = 1

cotθ =−

24 7

Sincecsc2_{θ= cot}2_{θ+ 1}

cscθ =−

q

−7 24

2

+ 1 =−25

24

and sinθ =−24

25

Sincesec2θ= 1 + tan2θ

secθ = q

−24 7

2

+ 1 = 25

7 and cosθ = 7 25

So, sin 2θ =2 cosθsinθ = 2 −24 25 7 25

Example: Ifcotθ =− 7

24 and θ ∈

−π

2,0

, find sin 2θ. Solution:

tanθ = 1

cotθ =−

24 7

Sincecsc2_{θ= cot}2_{θ+ 1}

cscθ =−

q

−7 24

2

+ 1 =−25

24 and sinθ =− 24 25

Sincesec2θ= 1 + tan2θ

secθ = q

−24 7

2

+ 1 = 25

7 and cosθ = 7 25

So, sin 2θ =2 cosθsinθ = 2 −24 25 7 25

Example: Ifcotθ =− 7

24 and θ ∈

−π

2,0

, find sin 2θ. Solution:

tanθ = 1

cotθ =−

24 7

Sincecsc2_{θ= cot}2_{θ+ 1}

cscθ =−

q

−7 24

2

+ 1 =−25

24 and sinθ =− 24 25

Sincesec2θ= 1 + tan2θ

secθ = q

−24 7

2

+ 1 = 25

7 and cosθ = 7 25

So, sin 2θ =2 cosθsinθ = 2 −24 25 7 25

Example: Ifcotθ =− 7

24 and θ ∈

−π

2,0

, find sin 2θ. Solution:

tanθ = 1

cotθ =−

24 7

Sincecsc2_{θ= cot}2_{θ+ 1}

cscθ =−

q

−7 24

2

+ 1 =−25

24 and sinθ =− 24 25

Sincesec2θ= 1 + tan2θ

secθ = q −24 7 2 + 1 = 25

7 and cosθ = 7 25

So, sin 2θ =2 cosθsinθ = 2 −24 25 7 25

Example: Ifcotθ =− 7

24 and θ ∈

−π

2,0

, find sin 2θ. Solution:

tanθ = 1

cotθ =−

24 7

Sincecsc2_{θ= cot}2_{θ+ 1}

cscθ =−

q

−7 24

2

+ 1 =−25

24 and sinθ =− 24 25

Sincesec2θ= 1 + tan2θ

secθ = q

−24 7

2

+ 1 = 25 7

and cosθ = 7 25

So, sin 2θ =2 cosθsinθ = 2 −24 25 7 25

Example: Ifcotθ =− 7

24 and θ ∈

−π

2,0

, find sin 2θ. Solution:

tanθ = 1

cotθ =−

24 7

Sincecsc2_{θ= cot}2_{θ+ 1}

cscθ =−

q

−7 24

2

+ 1 =−25

24 and sinθ =− 24 25

Sincesec2θ= 1 + tan2θ

secθ = q

−24 7

2

+ 1 = 25

7 and cosθ = 7 25

So, sin 2θ =2 cosθsinθ = 2 −24 25 7 25

Example: Ifcotθ =− 7

24 and θ ∈

−π

2,0

, find sin 2θ. Solution:

tanθ = 1

cotθ =−

24 7

Sincecsc2_{θ= cot}2_{θ+ 1}

cscθ =−

q

−7 24

2

+ 1 =−25

24 and sinθ =− 24 25

Sincesec2θ= 1 + tan2θ

secθ = q

−24 7

2

+ 1 = 25

7 and cosθ = 7 25

So, sin 2θ =2 cosθsinθ =

2 −24 25 7 25

Example: Ifcotθ =− 7

24 and θ ∈

−π

2,0

, find sin 2θ. Solution:

tanθ = 1

cotθ =−

24 7

Sincecsc2_{θ= cot}2_{θ+ 1}

cscθ =−

q

−7 24

2

+ 1 =−25

24 and sinθ =− 24 25

Sincesec2θ= 1 + tan2θ

secθ = q

−24 7

2

+ 1 = 25

7 and cosθ = 7 25

So, sin 2θ =2 cosθsinθ = 2 −24 25 7 25

Example: Evaluatesinπ 8cos

π

8.

Solution:

sinπ 8cos

π

8 =

1 2

2 sinπ

8cos

π

8

= 1

2sin

2· π

8

= 1

2·

√

2 2

=

√

Example: Evaluatesinπ 8cos

π

8.

Solution:

sinπ 8cos

π

8

= 1 2

2 sinπ

8cos

π

8

= 1

2sin

2· π

8

= 1

2·

√

2 2

=

√

Example: Evaluatesinπ 8cos

π

8.

Solution:

sinπ 8cos

π

8 =

1 2

2 sinπ

8cos

π

8

= 1

2sin

2· π

8

= 1

2·

√

2 2

=

√