Deriving Trig Identities

Precalculus / Calculus
Deriving Trig Identities
Identities [1]–[5] you memorize will be used to derive all remaining identities [8]–[24]. We must practice
the 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
Difference Identities
Pythagorean Identities
[1]
[2]
[3]
[4]
[5]
sin(u + v) = sin(u)cos(v) + cos(u)sin(v)
cos(u + v) = cos(u)cos(v) – sin(u)sin(v)
sin(u – v) = sin(u)cos(v) – cos(u)sin(v)
cos(u – v) = cos(u)cos(v) + sin(u)sin(v)
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+
sin 2 (!)
2
cos (!)
=
1
cos2 (!)
! sin(!) $ 2 ! 1 $ 2
& =#
&
1+ #
# cos(!) &
# cos(!) &
"
%
"
%
1 + tan2(˙) = sec2(˙)
[9]
Õ
cot2(˙) + 1 = csc2(˙)
How to Derive:
We start with cos2(˙) + sin2(˙) = 1. Divide both sides of this identity by sin2(˙). Then, we have
cos2 (!)
2
sin (!)
+1=
1
sin 2 (!)
! cos(!) $ 2
!
$2
#
& +1= # 1 &
# sin(!) &
# sin(!) &
"
%
"
%
2
cot (˙) + 1 = csc2(˙)
Prepared by Kunio Mitsuma, Ph.D.
Õ
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(˙)
Õ
[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(˙)
Õ
[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(˙) – sin2(˙)
cos(2˙) = cos2(˙) – [1 – cos2(˙)]
cos(2˙) = cos2(˙) – 1 + cos2(˙)
cos(2˙) = 2cos2(˙) – 1
Õ
[13] cos(2˙) = 1 – 2sin2(˙)
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(˙)
cos(2˙) = [1 – sin2(˙)] – sin2(˙)
cos(2˙) = 1 – sin2(˙) – sin2(˙)
cos(2˙) = 1 – 2sin2(˙)
Õ
Prepared by Kunio Mitsuma, Ph.D.
Power-Reduction Identities
1 + cos(2!)
[14] cos2(˙) =
2
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(˙)
cos(2˙) = cos2(˙) – [1 – cos2(˙)]
cos(2˙) = cos2(˙) – 1 + cos2(˙)
cos(2˙) = 2cos2(˙) – 1
1 + cos(2˙) = 2cos2(˙)
1 + cos(2!)
= cos2 (!)
2
1 + cos(2!)
cos2 (!) =
2
Õ
1! cos(2!)
2
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(˙)
cos(2˙) = [1 – sin2(˙)] – sin2(˙)
cos(2˙) = 1 – sin2(˙) – sin2(˙)
cos(2˙) = 1 – 2sin2(˙)
2sin2(˙) = 1 – cos(2˙)
[15] sin2(˙) =
sin 2 (!) =
1! cos(2!)
2
Õ
1! cos(2!)
1 + cos(2!)
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(˙)
2
cos(2˙) = [1 – sin (˙)] – 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(˙)
[16] tan 2 (!) =
sin 2 (!) =
1! cos(2!)
2
and
cos2 (!) =
1 + cos(2!)
2
Therefore,
1! cos(2!)
1! cos(2!)
2
tan 2 (!) =
=
=
2
cos (!) 1 + cos(2!) 1 + cos(2!)
2
sin 2 (!)
Prepared by Kunio Mitsuma, Ph.D.
Õ
Product-to-Sum Identities
[17] sin(u)cos(v) = 12 [sin(u + v) + sin(u – v)]
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) = 2 sin(u)cos(v)
sin(u + v)+ sin(u !v)
= sin(u)cos(v)
2
sin(u)cos(v) = 12 "$sin(u + v)+ sin(u !v)%'
#
&
Õ
[18] cos(u)sin(v) = 12 [sin(u + v) – sin(u – v)]
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)cos(v)+ cos(u)sin(v)
sin(u !v) = sin(u)cos(v)! cos(u)sin(v)
sin(u + v)! sin(u !v) = 2 cos(u)sin(v)
sin(u + v)! sin(u !v)
= cos(u)sin(v)
2
cos(u)sin(v) = 12 "$sin(u + v)! sin(u !v)%'
#
&
Õ
1
2
[19] cos(u)cos(v) = [cos(u – v) + cos(u + v)]
How to Derive:
We start with cos(u – v) = cos(u)cos(v) + sin(u)sin(v) and cos(u + v) = cos(u)cos(v) – sin(u)sin(v) .
Line them up and add both sides. Then,
cos(u !v) = cos(u)cos(v)+ sin(u)sin(v)
cos(u + v) = cos(u)cos(v)! sin(u)sin(v)
cos(u !v)+ cos(u + v) = 2 cos(u)cos(v)
cos(u !v)+ cos(u + v)
= cos(u)cos(v)
2
cos(u)cos(v) = 12 "$cos(u !v)+ cos(u + v)%'
#
&
Prepared by Kunio Mitsuma, Ph.D.
Õ
[20] sin(u)sin(v) = 12 [cos(u – v) – 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)cos(v)+ sin(u)sin(v)
cos(u + v) = cos(u)cos(v)! sin(u)sin(v)
cos(u !v)! cos(u + v) = 2 sin(u)sin(v)
cos(u !v)! cos(u + v)
= sin(u)sin(v)
2
sin(u)sin(v) = 12 "$cos(u !v)! cos(u + v)%'
#
&
Õ
Sum-to-Product Identities
! A + B $& ! A 'B $&
& cos ##
&
[21] sin(A) + sin(B) = 2 sin ###
#" 2 &&% ##" 2 &&%
How to Derive:
We start with sin(u + v) = sin(u)cos(v) + cos(u)sin(v) and sin(u – v) = sin(u)cos(v) – cos(u)sin(v).
Line them up and add both sides. Then,
sin(u + v) = sin(u)cos(v)+ cos(u)sin(v)
sin(u !v) = sin(u)cos(v)! cos(u)sin(v)
sin(u + v)+ sin(u !v) = 2 sin(u)cos(v)
sin(u + v)+ sin(u !v)
= sin(u)cos(v)
2
sin(u)cos(v) = 12 "$sin(u + v)+ sin(u !v)%'
#
&
Let A = u + v and B = u – v. Then,
A + B = 2u and A – B = 2v
A+B
A !B
u=
and v =
2
2
Substitution yields
! A + B $& ! A 'B $&
&& cos ##
&& = 12 (*sin(A)+ sin(B)+sin ###
)
,
"# 2 &% #"# 2 &%
! A + B $& ! A 'B $&
& cos ##
& = sin(A)+ sin(B)
2 sin ###
#" 2 &&% ##" 2 &&%
! A + B $& ! A 'B $&
& cos ##
&
sin(A)+ sin(B) = 2 sin ###
#" 2 &&% ##" 2 &&%
Õ
Prepared by Kunio Mitsuma, Ph.D.
! A + B $& ! A 'B $&
& sin ##
&
[22] sin(A) – sin(B) = 2 cos ###
#" 2 &&% ##" 2 &&%
How to Derive:
We start with sin(u + v) = sin(u)cos(v) + cos(u)sin(v) and sin(u – v) = sin(u)cos(v) – cos(u)sin(v).
Line them up and subtract both sides. Then,
sin(u + v) = sin(u)cos(v)+ cos(u)sin(v)
sin(u !v) = sin(u)cos(v)! cos(u)sin(v)
sin(u + v)! sin(u !v) = 2 cos(u)sin(v)
sin(u + v)! sin(u !v)
= cos(u)sin(v)
2
cos(u)sin(v) = 12 "$sin(u + v)! sin(u !v)%'
#
&
Let A = u + v and B = u – v. Then,
A + B = 2u and A – B = 2v
A+B
A !B
u=
and v =
2
2
Substitution yields
! A + B $& ! A 'B $&
& sin ##
& = 1 (sin(A)' sin(B)+cos ###
,
#" 2 &&% ##" 2 &&% 2 )*
! A + B $& ! A 'B $&
& sin ##
& = sin(A)' sin(B)
2 cos ###
#" 2 &&% ##" 2 &&%
! A + B $& ! A 'B $&
& sin ##
&
sin(A)' sin(B) = 2 cos ###
#" 2 &%& ##" 2 &&%
Õ
! A + B $& ! A 'B $&
& cos ##
&
[23] cos(A) + cos(B) = 2 cos ###
#" 2 &&% ##" 2 &&%
How to Derive:
We start with cos(u – v) = cos(u)cos(v) + sin(u)sin(v) and cos(u + v) = cos(u)cos(v) – sin(u)sin(v) .
Line them up and add both sides. Then,
cos(u !v) = cos(u)cos(v)+ sin(u)sin(v)
cos(u + v) = cos(u)cos(v)! sin(u)sin(v)
cos(u !v)+ cos(u + v) = 2 cos(u)cos(v)
cos(u !v)+ cos(u + v)
= cos(u)cos(v)
2
cos(u)cos(v) = 12 "$cos(u !v)+ cos(u + v)%'
#
&
Let A = u + v and B = u – v. Then,
A + B = 2u and A – B = 2v
A+B
A !B
u=
and v =
2
2
Substitution yields
Prepared by Kunio Mitsuma, Ph.D.
! A + B $& ! A 'B $&
& cos ##
& = 1 (cos(A)+ cos(B)+cos ###
,
#" 2 &&% ##" 2 &&% 2 *)
! A + B $& ! A 'B $&
& cos ##
& = cos(A)+ cos(B)
2 cos ###
#" 2 &&% ##" 2 &&%
! A + B $& ! A 'B $&
& cos ##
&
cos(A)+ cos(B) = 2 cos ###
#" 2 &&% ##" 2 &&%
Õ
! A + B $& ! A 'B $&
& sin ##
&
[24] cos(B) – cos(A) = 2 sin ###
#" 2 &&% ##" 2 &&%
How to Derive:
We start with cos(u – v) = cos(u)cos(v) + sin(u)sin(v) and cos(u + v) = cos(u)cos(v) – sin(u)sin(v) .
Line them up and subtract both sides. Then,
cos(u !v) = cos(u)cos(v)+ sin(u)sin(v)
cos(u + v) = cos(u)cos(v)! sin(u)sin(v)
cos(u !v)! cos(u + v) = 2 sin(u)sin(v)
cos(u !v)! cos(u + v)
= sin(u)sin(v)
2
sin(u)sin(v) = 12 "$cos(u !v)! cos(u + v)%'
#
&
Let A = u + v and B = u – v. Then,
A + B = 2u and A – B = 2v
A+B
A !B
u=
and v =
2
2
Substitution yields
! A + B $& ! A 'B $&
& sin ##
& = 1 (cos(B)' cos(A)+sin ###
,
#" 2 &&% ##" 2 &&% 2 *)
! A + B $& ! A 'B $&
& sin ##
& = cos(B)' cos(A)
2 sin ###
#" 2 &&% ##" 2 &&%
! A + B $& ! A 'B $&
& sin ##
&
cos(B)' cos(A) = 2 sin ###
#" 2 &&% ##" 2 &&%
Õ
Prepared by Kunio Mitsuma, Ph.D.