Transcript 5.4 PP

5.4
Multiple
Angle
Identities
Copyright © 2011 Pearson, Inc.
What you’ll learn about




Double-Angle Identities
Power-Reducing Identities
Half-Angle Identities
Solving Trigonometric Equations
… and why
These identities are useful in calculus courses.
Copyright © 2011 Pearson, Inc.
Slide 5.4 - 2
Double Angle Identities
sin 2u = 2 sin u cosu
ìcos 2 u - sin 2 u
ï
cos 2u = í2 cos 2 u - 1
ï1 - 2sin 2 u
î
2 tan u
tan 2u =
1 - tan 2 u
Copyright © 2011 Pearson, Inc.
Slide 5.4 - 3
Proving a Double-Angle Identity
cos 2x = cos(x + x)
= cos x cos x - sin x sin x
= cos 2 x - sin 2 x
Copyright © 2011 Pearson, Inc.
Slide 5.4 - 4
Power-Reducing Identities
1 - cos 2u
sin u =
2
1 + cos 2u
2
cos u =
2
1 - cos 2u
2
tan u =
1 + cos 2u
2
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Slide 5.4 - 5
Example Reducing a Power of 4
Rewrite sin 4 x in terms of trigonometric functions with
no power greater than 1.
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Slide 5.4 - 6
Example Reducing a Power of 4
(
sin x = sin x
4
2
)
2
æ 1 - cos 2x ö
=ç
÷ø
è
2
2
1 - 2 cos 2x + cos 2 2x
=
4
1 cos 2x 1 æ 1 + cos 4 x ö
= + ç
÷ø
4
2
4è
2
1 cos 2x 1 + cos 4 x
= +
4
2
8
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Slide 5.4 - 7
Half-Angle Identities
u
1 - cosu
sin = ±
2
2
u
1 + cosu
cos = ±
2
2
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ì 1 - cosu
ï±
1
+
cosu
ï
u ï1 - cosu
tan = í
2 ï sinu
ï sinu
ï1 + cosu
î
Slide 5.4 - 8
Example Using a Double Angle
Identity
Solve cos x + cos 3x = 0 in the interval [0,2p ).
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Slide 5.4 - 9
Example Using a Double Angle
Identity
Solve cos x + cos 3x = 0 in the interval [0,2p ).
Solve Graphically
The graph suggest that
there are six solutions:
0.79, 1.57, 2.36,
3.93, 4.71, 5.50.
Copyright © 2011 Pearson, Inc.
Slide 5.4 - 10
Example Using a Double Angle
Identity
Solve cos x + cos 3x = 0 in the interval [0,2p ).
Confirm Algebraically
cos x + cos 3x = 0
cos x + cos x cos 2x - sin x sin 2x = 0
(
)
cos x + cos x 1 - 2sin 2 x - sin x ( 2sin x cos x ) = 0
cos x + cos x - 2 cos x sin 2 x - 2 cos x sin 2 x = 0
2 cos x - 4 cos x sin 2 x = 0
(
)
2 cos x 1 - 2sin 2 x = 0
Copyright © 2011 Pearson, Inc.
Slide 5.4 - 11
Example Using a Double Angle
Identity
(
)
2 cos x 1 - 2sin 2 x = 0
cos x = 0
p
3p
x = or
2
2
p
3p
x = or
2
2
or
or
or
1 - 2 sin 2 x = 0
2
sin x = ±
2
p 3p 5p
7p
x = , , , or
4 4 4
4
The six exact solutions in the given interval are
p p
3p 5p 3p
7p
, ,
,
,
, and
.
4 2 4 4 2
4
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Slide 5.4 - 12
Quick Review
Find the general solution of the equation.
1. cot x + 1 = 0
2. (sin x)(1 - cos x) = 0
3. cos x - sin x = 0
(
)
4. 2sin x + 2 ( 2 sin x - 1) = 0
5. Find the height of the isosceles triangle with
base length 6 and leg length 4.
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Slide 5.4 - 13
Quick Review Solutions
Find the general solution of the equation.
1. cot x + 1 = 0
3p
x=
+ pn
4
2. (sin x)(1 - cos x) = 0
3. cos x - sin x = 0
Copyright © 2011 Pearson, Inc.
x = pn
x=
p
4
+ pn
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Quick Review Solutions
Find the general solution of the equation.
(
)
4. 2sin x + 2 ( 2 sin x - 1) = 0
5p
7p
x=
+ 2p n, x =
+ 2p n,
4
4
p
5p
x = + 2p n, x =
+ 2p n
6
6
5. Find the height of the isosceles triangle with
base length 6 and leg length 4.
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