急!!求二倍角公式,以及推导过程!
2个回答
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正弦二倍角公式: sin2α
=
2cosαsinα
推导:sin2α
=
sin(α+α)
=
sinαcosα
+
cosαsinα=
2sinαcosα
余弦二倍角公式: 余弦二倍角公式有三组表示形式,三组形式等价:
1.cos2α
=
2cos^2
α-
1
2.cos2α
=
1
−
2sin^2
α
3.cos2α
=
cos^2
α
−
sin^2
α
推导: cos2A
=
cos(A+A)
=
cosAcosA
-
sinAsinA
=
cos^2
A-
sin^2
A
=
2cos^2
A
-
1=1
-
2sin^2
A
正切二倍角公式:
tan2α
=
2tanα/[1
-
(tan^2α)]
tan(1/2*α)=(sin
α)/(1+cos
α)=(1-cos
α)/sin
α
推导: tan(2a)
=
tan(a+a)
=
(tan(a)
+
tan(a))/(1
-
tan(a)*tan(a)
)=
2tanα/[1
-
(tanα)^2]
降幂公式(半角公式): cos^2(A)=
[1
+
cos2A]/2
sin^2(A)=
[1
-
cos2A]/2
tan^2(A)=
[1-
cos2A]/[1+cos2A]
变式: sin2α
=
sin^2(α+π/4)
-
cos^2(α+π/4)
=
2sin^2(a+π/4)
-
1
=
1
-
2cos^2(α+π/4);
cos2α
=
2sin(α+π/4)cos(α+π/4)
=
2cosαsinα
推导:sin2α
=
sin(α+α)
=
sinαcosα
+
cosαsinα=
2sinαcosα
余弦二倍角公式: 余弦二倍角公式有三组表示形式,三组形式等价:
1.cos2α
=
2cos^2
α-
1
2.cos2α
=
1
−
2sin^2
α
3.cos2α
=
cos^2
α
−
sin^2
α
推导: cos2A
=
cos(A+A)
=
cosAcosA
-
sinAsinA
=
cos^2
A-
sin^2
A
=
2cos^2
A
-
1=1
-
2sin^2
A
正切二倍角公式:
tan2α
=
2tanα/[1
-
(tan^2α)]
tan(1/2*α)=(sin
α)/(1+cos
α)=(1-cos
α)/sin
α
推导: tan(2a)
=
tan(a+a)
=
(tan(a)
+
tan(a))/(1
-
tan(a)*tan(a)
)=
2tanα/[1
-
(tanα)^2]
降幂公式(半角公式): cos^2(A)=
[1
+
cos2A]/2
sin^2(A)=
[1
-
cos2A]/2
tan^2(A)=
[1-
cos2A]/[1+cos2A]
变式: sin2α
=
sin^2(α+π/4)
-
cos^2(α+π/4)
=
2sin^2(a+π/4)
-
1
=
1
-
2cos^2(α+π/4);
cos2α
=
2sin(α+π/4)cos(α+π/4)
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