Download Double Angle Identities: Derived and Applied

Double-Angle Identities
What s a double angle?
if _ θ _ is _ our _ angle....
2θ _ is _ a _ DOUBLE _ ANGLE
Like the addition and subtraction of angles, standard
thinking doesn t work on double angles.
let _ θ = 30
3
sin 2θ = sin 2(30 ) = sin 60 =
2

! 1$
2sin ! = 2sin 30 = 2 # & = 1
" 2%
!

sin 2θ ≠ 2 sin θ
Deriving the double angle formula for sine:
sin 2θ = sin(θ + θ )
Now use the addition identity
sin(θ + θ ) = sin θ cosθ + cosθ sin θ
sin 2θ = 2 sin θ cosθ
Deriving the double angle formula for cosine:
cos 2θ = cos(θ + θ )
Now use the addition identity
cos(θ + θ ) = cosθ cosθ − sin θ sin θ
cos 2θ = cos θ − sin θ
2
2
Re member:sin 2 θ + cos2 θ = 1, so...
cos2 θ = 1 − sin 2 θ
sin θ = 1 − cos θ
2
2
We have to more forms of the identity by using the
Pythagorean identity:
cos 2θ = cos θ − sin θ
2
2
Re member:sin 2 θ + cos2 θ = 1, so...
cos2 θ = 1 − sin 2 θ
sin 2 θ = 1 − cos2 θ
cos 2θ = cos2 θ − (1 − cos2 θ )
cos 2θ = 2 cos2 θ − 1
cos 2θ = (1 − sin 2 θ ) − sin 2 θ
cos 2θ = 1 − 2 sin 2 θ
Now, the identity for tangent:
tan u + tan v
tan(u + v ) =
1 − tan u tan v
tan θ + tan θ
tan(θ + θ ) =
1 − tan θ tan θ
2 tan θ
tan 2θ =
2
1 − tan θ
Using the Double Angle Identities:
12
sin θ = ,θ _ is_ in _1st _ Quadrant
13
Find _ the_ exact _ values_ of ....
1.sin 2θ =
finding _ cos! :
cos! = 1 ! sin 2 !
2
25
5
" 12 %
cos! = 1 ! $ ' =
=
# 13 &
169 13
sin 2! = 2sin ! cos!
! 12 $ ! 5 $ 120
sin 2! = 2 # & # & =
" 13 % " 13 % 169
Using the Double Angle Identities:
12
sin θ = ,θ _ is_ in _1st _ Quadrant
13
Find _ the_ exact _ values_ of ....
2.cos 2θ =
cos2! = cos2 ! ! sin 2 !
2
2
25 144 !119
" 5 % " 12 %
cos2! = $ ' ! $ ' =
!
=
# 13 & # 13 &
169 169 169
Using the Double Angle Identities:
12
sin θ = ,θ _ is_ in _1st _ Quadrant
13
Find _ the_ exact _ values_ of ....
3.tan 2θ =
2 tan !
tan 2! =
1 ! tan 2 !
12
24
(2)
!120
5
5
tan 2! =
=
2 =
!119 119
" 12 %
1! $ '
25
# 5&
sin θ
tan θ =
cosθ
12
12
13
tan θ =
=
5
5
13
Use the double angle identities to rewrite the following
expression as a single trigonometric function
cos 3t − sin 3t
2
2
Note it has the form of cosine double angle
identity
let _ θ = 3t
rewrite:
cos 2θ = cos2 θ − sin 2 θ
substitute _ out _ θ :
cos 2( 3t ) = cos2 3t − sin 2 3t
so...... = cos 6t
(
!1 " 8 % +
Find _ the _ exact _ value _ of _ sin * 2 cos $ ' # 17 & ,
)
Draw the triangle.
(
!1 " 8 % +
Find _ the _ exact _ value _ of _ sin * 2 cos $ ' # 17 & ,
)
(
!1 " 8 % +
sin * 2 cos $ ' - = sin [ 2! ]
# 17 & ,
)
sin [ 2! ] = 2sin ! cos!
15
17
θ
8
! 15 $ ! 8 $ 240
sin [ 2! ] = 2 # & # & =
" 17 % " 17 % 289
Verifying an identity using double angle identities:
Verify:
2 cos 2t
− 1 = cot t
2
sin 2t − 2 sin t
a lg ebra! :
(cost + sin t ) !1 = cot t
sin t
cost sin t
+
!1 = cot t
sin t sin t
use _ double _ angle _ identities :
(
2 cos2 t ! sin 2 t
)
2sin t cost ! 2sin t
2
!1 = cot t
factor :
2 ( cost ! sin t ) ( cost + sin t )
!1 = cot t
2sin t ( cost ! sin t )
simplify:
cos t
+ 1 − 1 = cot t
sin t
cos t
= cot t → YES
sin t