三角函数倍角公式推导

2023-12-27 10:52:24

1.sin(2α)=2sin(α)cos(α)

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已知单位圆中有圆周三角形△ABC,圆周角为α,圆心为点O
∴ 圆心角 = 2 ? 圆周角 ∴圆心角=2*圆周角 圆心角=2?圆周角
∵ ∠ A O C = α , A B = 2 c o s ( α ) , A C = 2 s i n ( α ) ∵∠AOC=α,AB=2cos(α),AC=2sin(α) AOC=α,AB=2cos(α),AC=2sin(α)
做 A P ⊥ B C 于点 P 做AP⊥BC于点P APBC于点P
∴ △ A B P 与△ A P O 共用边 A P ∴△ABP与△APO共用边AP ABPAPO共用边AP
∵ s i n ( α ) ? A B = s i n ( 2 α ) ∵sin(α)*AB=sin(2α) sin(α)?AB=sin(2α)
∵ s i n ( 2 α ) = 2 s i n ( α ) c o s ( α ) ∵sin(2α)=2sin(α)cos(α) sin(2α)=2sin(α)cos(α)

2. c o s ( 2 α ) = c o s ( α ) 2 ? s i n ( α ) 2 cos(2α)=cos(α)^2-sin(α)^2 cos(2α)=cos(α)2?sin(α)2

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∴ 圆心角 = 2 ? 圆周角 ∴圆心角=2*圆周角 圆心角=2?圆周角
∵ ∠ A O C = α , A B = 2 c o s ( α ) , A C = 2 s i n ( α ) ∵∠AOC=α,AB=2cos(α),AC=2sin(α) AOC=α,AB=2cos(α),AC=2sin(α)
做 A P ⊥ B C 于点 P 做AP⊥BC于点P APBC于点P
∵ ∠ P A C + ∠ A C P = ∠ A B C + ∠ A C P ∵∠PAC+∠ACP=∠ABC+∠ACP PAC+ACP=ABC+ACP
∵ ∠ P A C = α ∵∠PAC=α PAC=α
∵ O P = 1 ? 2 s i n ( α ) s i n ( α ) = c o s ( 2 α ) ∵OP=1-2sin(α)sin(α)=cos(2α) OP=1?2sin(α)sin(α)=cos(2α)
∵ c o s ( 2 α ) = 1 ? 2 s i n ( α ) 2 ∵cos(2α)=1-2sin(α)^2 cos(2α)=1?2sin(α)2
∵ c o s ( 2 α ) = c o s ( α ) 2 ? s i n ( α ) 2 ∵cos(2α)=cos(α)^2-sin(α)^2 cos(2α)=cos(α)2?sin(α)2

3. t a n ( 2 α ) = 2 t a n ( α ) 1 ? t a n ( α ) 2 tan(2α)=\frac{2tan(α)}{1-tan(α)^2} tan(2α)=1?tan(α)22tan(α)?

由公式1,2推tan(2α)
t a n ( 2 α ) = s i n ( 2 α ) c o s ( 2 α ) tan(2α)=\frac{sin(2α)}{cos(2α)} tan(2α)=cos(2α)sin(2α)?
= 2 s i n ( α ) c o s ( α ) c o s ( α ) 2 ? s i n ( α ) 2 =\frac{2sin(α)cos(α)}{cos(α)^2-sin(α)^2} =cos(α)2?sin(α)22sin(α)cos(α)?
= 2 t a n ( α ) 1 ? t a n ( α ) 2 =\frac{2tan(α)}{1-tan(α)^2} =1?tan(α)22tan(α)?

4. c o t ( 2 α ) = c o t ( α ) 2 ? 1 2 c o t ( α ) cot(2α)=\frac{cot(α)^2-1}{2cot(α)} cot(2α)=2cot(α)cot(α)2?1?

由公式1,2推cot(2α)
c o t ( 2 α ) = c o s ( 2 α ) s i n ( 2 α ) cot(2α)=\frac{cos(2α)}{sin(2α)} cot(2α)=sin(2α)cos(2α)?
= c o s ( α ) 2 ? s i n ( α ) 2 2 s i n ( α ) c o s ( α ) =\frac{cos(α)^2-sin(α)^2}{2sin(α)cos(α)} =2sin(α)cos(α)cos(α)2?sin(α)2?
= c o t ( α ) 2 ? 1 2 c o t ( α ) =\frac{cot(α)^2-1}{2cot(α)} =2cot(α)cot(α)2?1?

文章来源:https://blog.csdn.net/tyh751734196/article/details/135237320
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