Deriving Identities

8/8/2019 Deriving Identities

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Prepare d by Kunio Mitsuma, Ph.D.

Precalculus / Calculus

Deriving Trig Identities

Identities [1][5] you memorize will be used to derive all remaining identities [8][24]. We must practicethe process of deriving each of them until it becomes our second nature.

This handout has a complete process for each identity.

Identities to Memorize

Sum Identities [1] sin(u+ v) = sin(u)cos(v) + cos(u)sin(v)

[2] cos(u+ v) = cos(u)cos(v)sin(u)sin(v)

Difference Identities [3] sin(uv) = sin(u)cos(v) cos(u)sin(v)

[4] cos(uv) = cos(u)cos(v) + sin(u)sin(v)

Pythagorean Identities [5] cos2() + sin2() = 1

Pythagorean Identities (other two versions)

[8] 1 + tan2() = sec2()

How to Derive:

We start with cos2() + sin2() = 1. Divide both sides of this identity by cos2(). Then, we have

1+sin2(!)

cos2(!)=

1

cos2(!)

1+sin(!)

cos(!)

!

"

##

$

%

&&

2

=1

cos(!)

!

"

##

$

%

&&

2

1 + tan

2

() = sec

2

()

[9] cot2() + 1 = csc2()

How to Derive:

We start with cos2() + sin2() = 1. Divide both sides of this identity by sin2(). Then, we have

cos2(!)

sin2(!)+1=

1

sin2(!)

cos(!)

sin(!)

!

"

##

$

%

&&

2

+1=1

sin(!)

!

"

##

$

%

&&

2

cot2() + 1 = csc2()


2/7


Double-Angle Identities

[10] sin(2) = 2sin()cos()

How to Derive:

We start with sin(u + v) = sin(u)cos(v) + cos(u)sin(v). Let u= v = . Then,

sin(2) = sin()cos() + cos()sin()

sin(2) = sin()cos() + sin()cos()

sin(2) = 2sin()cos()

[11] cos(2) = cos2() sin2()

How to Derive:

We start with cos(u + v) = cos(u)cos(v) sin(u)sin(v). Let u= v = . Then,

cos(2) = cos()cos() sin()sin()

cos(2) = cos2() sin2()

[12] cos(2) = 2cos2() 1

How to Derive:

We start with cos(u + v) = cos(u)cos(v) sin(u)sin(v). Let u= v = . Then,cos(2) = cos()cos() sin()sin()


cos(2) = cos2() [1 cos2()]

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

cos(2) = 2cos2() 1

[13] cos(2) = 1 2sin2()

How to Derive:


cos(2) = cos()cos() sin()sin()cos(2) = cos2() sin2()

cos(2) = [1 sin2()] sin2()

cos(2) = 1 sin2() sin2()

cos(2) = 1 2sin2()


3/7


Power-Reduction Identities

[14] cos2() =1+ cos(2!)

2

How to Derive:




cos(2) = cos2() [1 cos2()]

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

cos(2) = 2cos2() 1

1 + cos(2) = 2cos2()

1+ cos(2!)

2= cos2(!)

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

2

[15] sin2() =1!cos(2!)

2

How to Derive:




cos(2) = [1 sin2()] sin2()

cos(2) = 1 sin2() sin2()

cos(2) = 1 2sin2()

2sin2() = 1 cos(2)

sin2(!)=1!cos(2!)

2

[16] tan2(!)=1!cos(2!)

1+ cos(2!)

How to Derive:




cos(2) = [1 sin2()] sin2() and cos(2) = cos2() [1 cos2()]

cos(2) = 1 sin2() sin2() and cos(2) = cos2() 1 + cos2()

cos(2) = 1 2sin2() and cos(2) = 2cos2() 1

2sin2() = 1 cos(2) and 1 + cos(2) = 2cos2()

sin2(!)=1!cos(2!)

2and cos2(!)=

1+ cos(2!)

2

Therefore,

tan2(!)=sin2(!)

cos2(!)=

1!cos(2!)

21+ cos(2!)

2

=1!cos(2!)

1+ cos(2!)


4/7


Product-to-Sum Identities

[17] sin(u)cos(v) = 12

[sin(u+ v) + sin(uv)]

How to Derive:

We start with sin(u + v) = sin(u)cos(v) + cos(u)sin(v) and sin(u v) = sin(u)cos(v) cos(u)sin(v).Line them up and add both sides. Then,

sin(u+v)= sin(u)cos(v)+ cos(u)sin(v)

sin(u!v)= sin(u)cos(v)!cos(u)sin(v)

sin(u+v)+ sin(u!v)= 2sin(u)cos(v)

sin(u+v)+ sin(u!v)

2= sin(u)cos(v)

sin(u)cos(v)= 12sin(u+v)+ sin(u!v)"#$

%&'

[18] cos(u)sin(v) = 12

[sin(u+ v) sin(uv)]

How to Derive:

We start with sin(u + v) = sin(u)cos(v) + cos(u)sin(v) and sin(u v) = sin(u)cos(v) cos(u)sin(v).

Line them up and subtract both sides. Then,



sin(u+v)!sin(u!v)= 2cos(u)sin(v)

sin(u+v)! sin(u!v)

2= cos(u)sin(v)

cos(u)sin(v)= 12sin(u+v)!sin(u!v)"#$

%&'

[19] cos(u)cos(v) = 12

[cos(uv) + cos(u+ v)]

How to Derive:

We start with cos(u v) = cos(u)cos(v) + sin(u)sin(v) and cos(u + v) = cos(u)cos(v) sin(u)sin(v) .Line them up and add both sides. Then,

cos(u!v)= cos(u)cos(v)+ sin(u)sin(v)

cos(u+v)= cos(u)cos(v)!sin(u)sin(v)

cos(u!v)+ cos(u+v)= 2cos(u)cos(v)

cos(u!v)+ cos(u+v)

2= cos(u)cos(v)

cos(u)cos(v)= 12cos(u!v)+ cos(u+v)"#$

%&'


5/7


[20] sin(u)sin(v) = 12

[cos(uv) cos(u+ v)]

How to Derive:

We start with cos(u v) = cos(u)cos(v) + sin(u)sin(v) and cos(u + v) = cos(u)cos(v) sin(u)sin(v) .Line them up and subtract both sides. Then,



cos(u!v)!cos(u+v)= 2sin(u)sin(v)

cos(u!v)!cos(u+v)

2= sin(u)sin(v)

sin(u)sin(v)= 12cos(u!v)!cos(u+v)"#$

%&'

Sum-to-Product Identities

[21] sin(A

) + sin(B

) =2sin

A+B

2

!

"

#

###

$

%

&&&&cos

A'B

2

!

"

#

###

$

%

&&&&

How to Derive:We start with sin(u + v) = sin(u)cos(v) + cos(u)sin(v) and sin(u v) = sin(u)cos(v) cos(u)sin(v).Line them up and add both sides. Then,



sin(u+v)+ sin(u!v)= 2sin(u)cos(v)

sin(u+v)+ sin(u!v)

2= sin(u)cos(v)

sin(u)cos(v)= 12sin(u+v)+ sin(u!v)"#$

%&'

Let A = u+ v and B= u v. Then,

A + B= 2u and A B= 2v

u=A+B

2and v=

A!B

2

Substitution yields

sinA+B

2

!

"

####

$

%

&&&&cos

A'B

2

!

"

####

$

%

&&&&= 1

2sin(A)+ sin(B)()*

+,-

2sinA+B

2

!

"

####

$

%

&&&&cos

A'B

2

!

"

####

$

%

&&&&= sin(A)+ sin(B)

sin(A)+ sin(B)= 2sinA+B

2

!

"####

$

%&&&&cos

A

'B

2

!

"####

$

%&&&&


6/7


[22] sin(A) sin(B) = 2cosA+B

2

!

"

####

$

%

&&&&sin

A'B

2

!

"

####

$

%

&&&&

How to Derive:We start with sin(u + v) = sin(u)cos(v) + cos(u)sin(v) and sin(u v) = sin(u)cos(v) cos(u)sin(v).Line them up and subtract both sides. Then,



sin(u+v)!sin(u!v)= 2cos(u)sin(v)

sin(u+v)! sin(u!v)

2= cos(u)sin(v)

cos(u)sin(v)= 12sin(u+v)!sin(u!v)"#$

%&'



u=A+B

2

and v=A!B

2

Substitution yields

cosA+B

2

!

"

####

$

%

&&&&sin

A'B

2

!

"

####

$

%

&&&&= 1

2sin(A)'sin(B)()*

+,-

2cosA+B

2

!

"

####

$

%

&&&&sin

A'B

2

!

"

####

$

%

&&&&= sin(A)'sin(B)

sin(A)'sin(B)= 2cosA+B

2

!

"####

$

%

&&&&sin

A'B

2

!

"####

$

%

&&&&

[23] cos(A) + cos(B) = 2cosA+B

2

!

"####

$

%&&&&cosA'B

2

!

"####

$

%&&&&

How to Derive:We start with cos(u v) = cos(u)cos(v) + sin(u)sin(v) and cos(u + v) = cos(u)cos(v) sin(u)sin(v) .Line them up and add both sides. Then,



cos(u!v)+ cos(u+v)= 2cos(u)cos(v)

cos(u!v)+ cos(u+v)

2= cos(u)cos(v)

cos(u)cos(v)= 1

2

cos(u!v)+ cos(u+v)"#$

%&'



u=A+B

2and v=

A!B

2

Substitution yields


7/7


cosA+B

2

!

"

####

$

%

&&&&cos

A'B

2

!

"

####

$

%

&&&&= 1

2cos(A)+ cos(B)()*

+,-

2cosA+B

2

!

"####

$

%

&&&&cos

A'B

2

!

"####

$

%

&&&&= cos(A)+ cos(B)

cos(A)+ cos(

B)= 2cos

A+B

2

!

"

####

$

%

&&&&cos

A'B

2

!

"

####

$

%

&&&&

[24] cos(B) cos(A) = 2sinA+B

2

!

"

####

$

%

&&&&sin

A'B

2

!

"

####

$

%

&&&&

How to Derive:We start with cos(u v) = cos(u)cos(v) + sin(u)sin(v) and cos(u + v) = cos(u)cos(v) sin(u)sin(v) .Line them up and subtract both sides. Then,



cos(u!v)!cos(u+v)= 2sin(u)sin(v)

cos(u!v)!cos(u+v)

2= sin(u)sin(v)

sin(u)sin(v)= 12cos(u!v)!cos(u+v)"#$

%&'



u=A+B

2and v=

A!B

2

Substitution yields

sinA+B

2

!

"####

$

%&&&&sin

A'B

2

!

"####

$

%&&&&= 1

2 cos(B)'cos(A)()* +,-

2sinA+B

2

!

"####

$

%

&&&&sin

A'B

2

!

"####

$

%

&&&&= cos(B)'cos(A)

cos(B)'cos(A)= 2sinA+B

2

!

"

####

$

%

&&&&sin

A'B

2

!

"

####

$

%

&&&&

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Deriving Identities