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Trigonometry
Notes on the Double and Half Angle Identities.
The Derivations of the Double and Half Angle Identities: To find the double angle identities, we just use the summation identities with both angles equal to A.
sin ( 2A ) = sin ( A + A ) = sin ( A ) cos ( A ) + cos ( A ) sin ( A )
= 2sin ( A ) cos ( A )
tan ( 2A ) =
=
tan ( A ) + tan ( A )
1 − tan ( A ) tan ( A )
2 tan ( A )
1 − tan 2 ( A )
cos ( 2A ) = cos ( A + A ) = cos ( A ) cos ( A ) − sin ( A ) sin ( A )
= cos 2 ( A ) − sin 2 ( A )
But, note on the last one, we could now use the Pythagorean identities on it.
cos ( 2A ) = cos 2 ( A ) − sin 2 ( A ) = (1 − sin 2 ( A ) ) − sin 2 ( A )
= 1 − 2sin 2 ( A )
cos ( 2A ) = cos 2 ( A ) − sin 2 ( A ) = cos 2 ( A ) − (1 − cos 2 ( A ) ) = cos 2 ( A ) − 1 + cos 2 ( A )
= 2 cos 2 ( A ) − 1
Thus, there are 3 double angle identities for cosine.
For the half angle identities, for sine and cosine, we take the last 2 double angle identities and solve for either sin ( A ) or cos ( A ) .
Then, we will make a substitution. Since, we will be replacing the variable, let’s start with the identities in x instead of A.
cos ( 2x ) = 1 − 2sin 2 ( x )
cos ( 2x ) = 2 cos 2 ( x ) − 1
2sin 2 ( x ) = 1 − cos ( 2x )
sin 2 ( x ) =
1 + cos ( 2x ) = 2 cos 2 ( x )
1− cos ( 2x )
2
sin ( x ) = ±
⎛A⎞
sin ⎜ ⎟ = ±
⎝2⎠
1− cos ( 2x )
2
±
1− cos ( A )
2
making A = 2x.
±
1+ cos ( 2x )
2
= cos 2 ( x )
1+ cos ( 2x )
2
= cos ( x )
1+ cos ( A )
2
⎛A⎞
= cos ⎜ ⎟
⎝2⎠
making A = 2x.
For the half angle identity for tangent, start by using the half angle identities above.
⎛A⎞
tan ⎜ ⎟ =
⎝2⎠
⎛A⎞
sin ⎜ ⎟
⎝2⎠
⎛A⎞
cos ⎜ ⎟
⎝2⎠
But, we can also make use of the double identities.
tan ( A2 ) =
( A2 )
=
cos ( A2 )
sin
=
±
±
1− cos ( A )
2
1+ cos ( A )
=±
2
1 − cos ( A )
1 + cos ( A )
.
sin ( 2( A2 ) )
( A2 ) cos( A2 )
sin ( A )
=
= 1+ cos ( A )
2 A
2 cos ( 2 )
1+ cos ( 2 ( A2 ) )
2sin
and
tan (
SCC:Rickman
A
2
( A )⎤⎦
1− cos ( A )
= sin ( A )
) = 1+ cos( A ) = 1+ cos( A ) 1− cos( A ) = sin (1A−)cos⎡⎣1− cos( A()A )⎤⎦ = sin ( Asin) ⎡⎣1−(cos
A)
sin ( A )
sin ( A )
1− cos ( A )
2
2
Notes on the Double and Half Angle Identities.
Page #1 of 4
The Trigonometric Double and Half Angle Identities: Thus, in summary,
The Trigonometric Double Angle Identities
The Trigonometric Half Angle Identities
cos ( 2A ) = cos ( A ) − sin ( A )
1 + cos ( A )
⎛A⎞
cos ⎜ ⎟ = ±
2
⎝2⎠
2
2
= 1 − 2sin 2 ( A )
1 − cos ( A )
⎛A⎞
sin ⎜ ⎟ = ±
2
⎝2⎠
= 2 cos 2 ( A ) − 1
sin ( 2A ) = 2sin ( A ) cos ( A )
tan ( 2A ) =
1 − cos ( A )
⎛A⎞
tan ⎜ ⎟ = ±
1 + cos ( A )
⎝2⎠
2 tan ( A )
1 − tan 2 ( A )
=
=
sin ( A )
1 + cos ( A )
1 − cos ( A )
sin ( A )
Example #1: Given cos ( A ) = - 23 , and A is in the 4th quadrant, evaluate a) cos ( A2 ) , b) sin ( A2 ) , c) tan ( A2 ) , d) sec ( A2 ) , e) cos ( 2A ) ,
f) sin ( 2A ) , and g) tan ( 2A )
►
First, find the quadrant for
A
2
cos ( A2 ) = ±
270° < A < 360°
135° <
Thus,
A
2
A
2
b)
a)
.
< 180°
nd
is in the 2 quadrant.
sin ( A2 ) = ±
1− cos ( A )
2
=-
1+ ( − 2 3)
2
=+
1− ( − 2 3)
2
=-
3− 2
6
=-
6
6
The selection of + or – was done based on the fact that
= 2 ( - 23 ) − 1
=
1
6
=-
e)
cos ( 2A ) = 2 cos 2 ( A ) − 1
c)
1+ cos ( A )
2
3+ 2
6
30
6
is in the 2nd quadrant.
A
2
sin 2 ( A ) = 1 − cos 2 ( A )
= 1 − ( - 32 ) = 99 − 54 =
2
= 89 − 1 = 89 − 99
1
=9
tan ( A ) =
sin ( A )
=
cos ( A )
5 9
-2 3
5
6
=-
1+ cos ( A )
=-
1− ( − 2 3)
=-
3+ 2
3− 2
1
cos A2
( )
=- 6
1+ ( − 2 3)
=-
5
1
=- 5
=
For parts f & g, we need sin(A) and tan(A) .
2
=
tan ( A2 ) = ±
d)
sec ( A2 ) =
1− cos ( A )
f)
sin ( 2A ) = 2sin ( A ) cos ( A )
g)
= 2 ( 95 )( - 32 )
5
9
2 tan ( A )
tan ( 2A ) = 1− tan 2 ( A )
=
20
=27
5
6
2 ( -5 6 )
1− ( -5 6 )
2
( -5 3)
= 1− ( 25
=
36 )
-60
36 − 25
=-
60
11
□
Example #2: Evaluate tan (15° ) using all 3 half angle identities for tangent.
►
tan (15° ) = ±
1− cos ( 30° )
1+ cos ( 30° )
=+
(2− 3 ) (2− 3 )
= 2+ 3 2− 3 =
( )( )
1−
1+
(
(
3 2
3 2
)
=
)
4− 4 3 +3
4−3
2− 3
2+ 3
sin ( 30° )
tan (15° ) = 1+ cos (30°) = 1+1 32 2 =
=
(
1
2+ 3
( 2− 3 )
=
) ( 2− 3 )
1
2+ 3
tan (15° ) =
=
2− 3
4 −3
1− cos ( 30° )
sin ( 30° )
=
1−
(
3 2
)
12
2− 3
1
= 2− 3
= 2− 3
= 7−4 3
(
Even though we got 2 different forms of the answer, they are actually equivalent, since 2 − 3
)
2
= 4−4 3+3= 7−4 3 .
□
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Notes on the Double and Half Angle Identities.
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Proving Identities:
Example #3: Prove cos ( 2x ) − cos ( x ) = -2sin 2 ( x2 ) ( 2 cos ( x ) + 1) .
►
cos ( 2x ) − cos ( x ) -2sin 2 ( x2 ) ( 2 cos ( x ) + 1)
2 cos 2 ( x ) − 1 − cos ( x )
⎛ 1 − cos ( x ) ⎞
-2 ⎜
⎟ ( 2 cos ( x ) + 1)
2
2 cos ( x ) − cos ( x ) − 1
⎝
⎠
( cos ( x ) − 1) ( 2 cos ( x ) + 1)
2
2 cos 2 ( x ) + cos ( x ) − 2 cos ( x ) − 1
2 cos 2 ( x ) − cos ( x ) − 1 = 2 cos 2 ( x ) − cos ( x ) − 1
□
Example #4: Prove tan ( 2x ) cos ( x ) ⎡⎣ 2 − sec 2 ( x ) ⎤⎦ = 2 cot ( x2 ) ⎡⎣1 − cos ( x ) ⎤⎦ .
►
tan ( 2x ) cos ( x ) ⎡⎣ 2 − sec2 ( x ) ⎤⎦ 2 cot ( x ) ⎡1 − cos ( x ) ⎤
2 ⎣
⎦
⎡ 2 tan2( x ) ⎤ cos ( x ) ⎡ 2 − (1 + tan 2 ( x ) ) ⎤
⎣
⎦ 2 ⎡⎢ tan1( x ) ⎤⎥ ⎡⎣1 − cos ( x ) ⎤⎦
⎣⎢ 1− tan ( x ) ⎦⎥
⎣ 2 ⎦
⎡ 2 tan2( x ) ⎤ cos ( x ) ⎡ 2 − 1 − tan 2 ( x ) ⎤
⎡
⎣
⎦ 2 1− cos x1 sin x ⎤ ⎡⎣1 − cos ( x ) ⎤⎦
⎢⎣ 1− tan ( x ) ⎥⎦
⎢⎣ ( ( )) ( ) ⎥⎦
⎡ 2 tan2( x ) ⎤ cos ( x ) ⎡1 − tan 2 ( x ) ⎤
sin ( x )
⎣
⎦ 2 ⎡ 1− cos ( x ) ⎤ ⎡⎣1 − cos ( x ) ⎤⎦
⎢⎣ 1− tan ( x ) ⎥⎦
⎣
⎦
2 tan ( x ) cos ( x ) 2sin ( x )
2
( ) cos ( x )
sin ( x )
cos ( x )
2sin ( x ) = 2sin ( x )
□
Solving Equations: Example #5: Find all real solutions of sin ( 2x ) + cos ( x ) = 0 for x in [ 0°,360° ) .
►
sin ( 2x ) + cos ( x ) = 0
2sin ( x ) cos ( x ) + cos ( x ) = 0
cos ( x ) ⎡⎣ 2sin ( x ) + 1⎤⎦ = 0
cos ( x ) = 0
or
π 3π
x= ,
2 2
2sin ( x ) + 1 = 0
2sin ( x ) = -1
-1
2
7 π 11π
x=
,
6 6
sin ( x ) =
x=
π 3π 7 π 11π
, , ,
2 2 6 6
□
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Notes on the Double and Half Angle Identities.
Page #3 of 4
Example #6: Find all real solutions of 2 cos 2 ( x2 ) + cos ( x ) = 0 for x in [ 0, 2π ) .
►
2 cos 2 ( x2 ) + cos ( x ) = 0
⎛ 1 + cos ( x ) ⎞
2⎜
⎟ + cos ( x ) = 0
2
⎝
⎠
⎛ 1 + cos ( x ) ⎞
2⎜
⎟ + cos ( x ) = 0
2
⎝
⎠
1 + cos ( x ) + cos ( x ) = 0
2 cos ( x ) = -1
cos ( x ) = - 12
x=
2π 4π
,
3 3
□
Applications: Example #7: If a spray can sprays with an angle of θ at a distance of d from a surface, show that the formula for the area of the
sprayed region is given by a) Area = π d 2 tan 2 ( θ2 ) and also b) Area =
π d 2 ⎡⎣1− cos ( θ ) ⎤⎦
1+ cos ( θ )
. Assume the can is being held parallel to the surface.
Plus, c) find the area of the spray when the angle of spray is 45˚ at a distance of 30cm.
►
A cross section of the cone forms an isosceles triangle, which can be split into 2 right triangles. See figure. This leads to the
following.
r
b)
c)
a)
= tan ( θ2 )
d
2
2
2
2 θ
π ( 30cm ) ⎡⎣1− cos ( 45° ) ⎤⎦
Area = π r
Area = π d tan ( 2 )
Area =
r = d tan ( θ2 )
1+ cos ( 45° )
2
2
= π ( d tan ( θ2 ) )
1− cos ( θ )
900 π ⎡1− 2 2 ⎤
= π d 2 ⎡ ± 1+ cos ( θ) ⎤
= 1+⎣ 2 2 ⎦ cm 2
⎢⎣
⎥⎦
2
2 θ
= π d tan ( 2 )
1− cos ( θ )
900 π ⎡ 2 − 2 ⎤
= π d 2 1+ cos( θ)
= 2 +⎣ 2 ⎦ cm 2
(
=
π d 2 ⎡⎣1− cos ( θ ) ⎤⎦
1+ cos ( θ )
)
=
900 π ⎡⎣ 4 − 4 2 + 2 ⎤⎦
4+ 2
cm 2
= 150π ⎡⎣6 − 4 2 ⎤⎦ cm 2
= 300π ⎡⎣3 − 2 2 ⎤⎦ cm 2
□
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Notes on the Double and Half Angle Identities.
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