一.几何推理
1.两角和公式
做一斜边为1的直角△ABC,任意旋转非
k
Π
,
k
=
N
kΠ,k=N
kΠ,k=N,补充如图,令
∠
A
B
C
=
∠
α
，
∠
C
B
F
=
∠
β
∠ABC=∠α，∠CBF=∠β
∠ABC=∠α，∠CBF=∠β
∴
∠
D
B
F
=
∠
D
B
A
+
∠
α
+
∠
β
=
90
°
,
∠
D
A
F
=
∠
D
B
A
+
∠
D
A
B
∴∠DBF=∠DBA+∠α+∠β=90°,∠DAF=∠DBA+∠DAB
∴∠DBF=∠DBA+∠α+∠β=90°,∠DAF=∠DBA+∠DAB
∵
∠
D
A
B
=
∠
α
+
∠
β
∵∠DAB=∠α+∠β
∵∠DAB=∠α+∠β
∴
∠
A
C
F
+
∠
B
C
F
=
90
°
∴∠ACF+∠BCF=90°
∴∠ACF+∠BCF=90°
∵
∠
A
C
F
=
∠
β
∵∠ACF=∠β
∵∠ACF=∠β
∴
A
B
长度为
1
∴AB长度为1
∴AB长度为1
∵
A
C
=
s
i
n
(
α
)
,
B
C
=
c
o
s
(
α
)
∵AC=sin(α),BC=cos(α)
∵AC=sin(α),BC=cos(α)
∵
B
F
=
c
o
s
(
α
)
∗
c
o
s
(
β
)
,
C
F
=
c
o
s
(
α
)
∗
s
i
n
(
β
)
,
A
E
=
s
i
n
(
α
)
s
i
n
(
β
)
,
C
E
=
s
i
n
(
α
)
c
o
s
(
β
)
,
B
D
=
E
F
=
s
i
n
(
α
+
β
)
,
D
A
=
c
o
s
(
α
+
β
)
∵BF=cos(α)*cos(β),CF=cos(α)*sin(β),AE=sin(α)sin(β),CE=sin(α)cos(β),BD=EF=sin(α+β),DA=cos(α+β)
∵BF=cos(α)∗cos(β),CF=cos(α)∗sin(β),AE=sin(α)sin(β),CE=sin(α)cos(β),BD=EF=sin(α+β),DA=cos(α+β)
∵
{
c
o
s
(
α
+
β
)
=
c
o
s
(
α
)
∗
c
o
s
(
β
)
−
s
i
n
(
α
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∗
s
i
n
(
β
)
s
i
n
(
α
+
β
)
=
s
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(
α
)
∗
c
o
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+
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∗
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∵\begin{cases} cos(α+β)=cos(α)*cos(β)-sin(α)*sin(β) \\sin(α+β)=sin(α)*cos(β)+cos(α)*sin(β) \end{cases}
∵{cos(α+β)=cos(α)∗cos(β)−sin(α)∗sin(β)sin(α+β)=sin(α)∗cos(β)+cos(α)∗sin(β)
2.两角差公式
∵
{
c
o
s
(
α
+
β
)
=
c
o
s
(
α
)
∗
c
o
s
(
β
)
−
s
i
n
(
α
)
∗
s
i
n
(
β
)
s
i
n
(
α
+
β
)
=
s
i
n
(
α
)
∗
c
o
s
(
β
)
+
c
o
s
(
α
)
∗
s
i
n
(
β
)
∵\begin{cases} cos(α+β)=cos(α)*cos(β)-sin(α)*sin(β) \\sin(α+β)=sin(α)*cos(β)+cos(α)*sin(β) \end{cases}
∵{cos(α+β)=cos(α)∗cos(β)−sin(α)∗sin(β)sin(α+β)=sin(α)∗cos(β)+cos(α)∗sin(β)
对
∠
β
做取反变化
对∠β做取反变化
对∠β做取反变化
∵
{
c
o
s
(
α
+
(
−
β
)
)
=
c
o
s
(
α
)
∗
c
o
s
(
β
)
−
s
i
n
(
α
)
∗
(
−
s
i
n
(
β
)
)
s
i
n
(
α
+
(
−
β
)
)
=
s
i
n
(
α
)
∗
c
o
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(
β
)
+
c
o
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(
α
)
∗
(
−
s
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β
)
)
∵\begin{cases} cos(α+(-β))=cos(α)*cos(β)-sin(α)*(-sin(β)) \\sin(α+(-β))=sin(α)*cos(β)+cos(α)*(-sin(β)) \end{cases}
∵{cos(α+(−β))=cos(α)∗cos(β)−sin(α)∗(−sin(β))sin(α+(−β))=sin(α)∗cos(β)+cos(α)∗(−sin(β))
∵ { c o s ( α − β ) = s i n ( α ) ∗ s i n ( β ) + c o s ( α ) ∗ s i n ( β ) s i n ( α − β ) = s i n ( α ) ∗ c o s ( β ) − c o s ( α ) ∗ s i n ( β ) ∵\begin{cases} cos(α-β)=sin(α)*sin(β)+cos(α)*sin(β) \\sin(α-β)=sin(α)*cos(β)-cos(α)*sin(β) \end{cases} ∵{cos(α−β)=sin(α)∗sin(β)+cos(α)∗sin(β)sin(α−β)=sin(α)∗cos(β)−cos(α)∗sin(β)
3.总结
∵ { c o s ( α + β ) = c o s ( α ) ∗ c o s ( β ) − s i n ( α ) ∗ s i n ( β ) s i n ( α + β ) = s i n ( α ) ∗ c o s ( β ) + c o s ( α ) ∗ s i n ( β ) c o s ( α − β ) = s i n ( α ) ∗ s i n ( β ) + c o s ( α ) ∗ s i n ( β ) s i n ( α − β ) = s i n ( α ) ∗ c o s ( β ) − c o s ( α ) ∗ s i n ( β ) ∵\begin{cases} cos(α+β)=cos(α)*cos(β)-sin(α)*sin(β) \\sin(α+β)=sin(α)*cos(β)+cos(α)*sin(β) \\cos(α-β)=sin(α)*sin(β)+cos(α)*sin(β) \\sin(α-β)=sin(α)*cos(β)-cos(α)*sin(β) \end{cases} ∵⎩ ⎨ ⎧cos(α+β)=cos(α)∗cos(β)−sin(α)∗sin(β)sin(α+β)=sin(α)∗cos(β)+cos(α)∗sin(β)cos(α−β)=sin(α)∗sin(β)+cos(α)∗sin(β)sin(α−β)=sin(α)∗cos(β)−cos(α)∗sin(β)
4.其他
为什么几何推理∠β和∠α不是钝角,根据诱导公式可将钝角化为锐角。所以只推导锐角和可以等价于推导任意角和
