9.3 and 9.4 The Spatial Model And Spatial Prediction and the Kriging Paradigm.
9.3 Double-Angle and Half-Angle Formulas -FURTHER IDENTITIES
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9.3 Double-Angle and Half- Angle Formulas -FURTHER IDENTITIES
description
9.3 Double-Angle and Half-Angle Formulas -FURTHER IDENTITIES. Double Angle Formulas. Suppose we use the sum formula for sine, but the two values that we add are the same, that is α = β. Double angle formula for cosine. Alternate forms of the cosine double angle formula. - PowerPoint PPT Presentation
Transcript of 9.3 Double-Angle and Half-Angle Formulas -FURTHER IDENTITIES
9.3 Double-Angle and Half-Angle Formulas
-FURTHER IDENTITIES
• Suppose we use the sum formula for sine, but the two values that we add are the same, that is α=β
Double Angle Formulas
sin( ) sin cos cos sin sin cos sin cos 2sin cos sin(2 )
Double angle formula for cosine
cos( ) cos cos sin sin
cos( ) cos cos sin sin
2 2cos(2 ) cos sin
Alternate forms of the cosine double angle formula
Because we know that sin2x+cos2x=1, we can alter that equation and solve for sin2x or cos2x and make
substitutions into the double angle formula.
2 2cos(2 ) cos sin
2 2cos(2 ) (1 sin ) sin 2cos(2 ) 1 2sin
• Now do the same thing but lets substitute for sin2x instead.
2 2cos(2 ) cos sin 2 2cos (1 cos )
2cos(2 ) 2cos 1
Double-angle Formula for Tangent
tantan
tan2
2
1 2
tantan1
tantantan2tan
Hwk, pg. 383 1-16, 19, 20
Copyright © 2011 Pearson Education, Inc. Slide 9.3-7
9.3 Finding Function Values of 2
Example Given and sin < 0, find sin 2,
cos 2, and tan 2.
Solution To find sin 2, we must find sin .
53cos
2524
53
54
22sin
cossin22sin
54
sin153
sin
1cossin2
2
22
Choose the negative square root since sin < 0.
Copyright © 2011 Pearson Education, Inc. Slide 9.3-8
9.3 Finding Function Values of 2
724
2cos2sin
2tan
or724
1
2
34
cossin
tan where,tan1tan2
2tan
257
54
53
sincos2cos
257
2524
234
34
53
54
2
22
22
Copyright © 2011 Pearson Education, Inc. Slide 9.3-9
9.3 Simplifying Expressions Using Double-Number Identities
Example Simplify each expression.
(a) cos² 7x – sin² 7x (b) sin 15° cos 15°
Solution
(b) cos 2A = cos² A – sin² A. Substituting 7x in for A gives cos² 7x – sin² 7x = cos 2(7x) = cos 14x.
(c) Apply sin 2A = 2 sin A cos A directly.
41
30sin21
)152sin(21
15cos15sin)2(21
15cos15sin
Copyright © 2011 Pearson Education, Inc. Slide 9.3-10
9.3 Half-Number Identities
• Half-number or half-angle identities for sine and cosine are used in calculus when eliminating the xy-term from an equation of the form Ax² + Bxy + Cy² + Dx + Ey + F = 0, so the type of conic it represents can be determined.
• From the alternative forms of the identity for cos 2A, we can derive three additional identities, e.g.
2cos1
2sin
.2
that so 2Let 2
2cos1sin
2cos1sin2
sin212cos2
2
AA
AxAx
xx
xx
xx
.sin2A
Choose the sign ± depending on the quadrant of the angle A/2.
Copyright © 2011 Pearson Education, Inc. Slide 9.3-11
9.3 Half-Number Identities
Half-Number Identities
2cos1
2sin
2cos1
2cos
AAAA
AAA
AAA
AAA
sincos1
2tan
cos1sin
2tan
cos1cos1
2tan
Copyright © 2011 Pearson Education, Inc. Slide 9.3-12
9.3 Using a Half-Number Identity to Find an Exact Value
Example Find the exact value of
Solution
.12
cos
232
22
223
1
223
1
26
cos1
26cos
12cos
Copyright © 2011 Pearson Education, Inc. Slide 9.3-13
9.3 Finding Function Values of x/2
Example Given
Solution The half-angle terminates in quadrant II since
.tan and ,sin,cos 222xxx
find ,2 with ,cos 23
32 xx
55
cossin
2tan
630
65
21
2cos
66
61
21
2sin
630
66
2
2
32
32
x
xx
x
x
.2 243
23 xx
Copyright © 2011 Pearson Education, Inc. Slide 9.3-14
9.3 Simplifying Expressions Using Half-Number Identities
Example Simplify the expression
Solution This matches the part of the identity for cos A/2. Replace A with 12x to get
.2
12cos1 x
.6cos2
12cos
212cos1
x
xx
• HWK pg. 383 1-12
• HWK pg. 384 19-22, 31-36
