三角函数的正确计算方法

公式,例题
2025-03-15 08:55:40
推荐回答(2个)
回答1:

倒数关系: 商的关系: 平方关系:
tanα ·cotα=1
sinα ·cscα=1
cosα ·secα=1 sinα/cosα=tanα=secα/cscα
cosα/sinα=cotα=cscα/secα sin2α+cos2α=1
1+tan2α=sec2α
1+cot2α=csc2α

诱导公式
sin(-α)=-sinα
cos(-α)=cosα tan(-α)=-tanα
cot(-α)=-cotα

sin(π/2-α)=cosα
cos(π/2-α)=sinα
tan(π/2-α)=cotα
cot(π/2-α)=tanα

sin(π/2+α)=cosα
cos(π/2+α)=-sinα
tan(π/2+α)=-cotα
cot(π/2+α)=-tanα

sin(π-α)=sinα
cos(π-α)=-cosα
tan(π-α)=-tanα
cot(π-α)=-cotα

sin(π+α)=-sinα
cos(π+α)=-cosα
tan(π+α)=tanα
cot(π+α)=cotα

sin(3π/2-α)=-cosα
cos(3π/2-α)=-sinα
tan(3π/2-α)=cotα
cot(3π/2-α)=tanα

sin(3π/2+α)=-cosα
cos(3π/2+α)=sinα
tan(3π/2+α)=-cotα
cot(3π/2+α)=-tanα

sin(2π-α)=-sinα
cos(2π-α)=cosα
tan(2π-α)=-tanα
cot(2π-α)=-cotα

sin(2kπ+α)=sinα
cos(2kπ+α)=cosα
tan(2kπ+α)=tanα
cot(2kπ+α)=cotα
(其中k∈Z)

两角和与差的三角函数公式 万能公式
sin(α+β)=sinαcosβ+cosαsinβ
sin(α-β)=sinαcosβ-cosαsinβ
cos(α+β)=cosαcosβ-sinαsinβ
cos(α-β)=cosαcosβ+sinαsinβ

tanα+tanβ
tan(α+β)=——————
1-tanα ·tanβ

tanα-tanβ
tan(α-β)=——————
1+tanα ·tanβ
2tan(α/2)
sinα=——————
1+tan2(α/2)

1-tan2(α/2)
cosα=——————
1+tan2(α/2)

2tan(α/2)
tanα=——————
1-tan2(α/2)

半角的正弦、余弦和正切公式 三角函数的降幂公式

二倍角的正弦、余弦和正切公式 三倍角的正弦、余弦和正切公式
sin2α=2sinαcosα

cos2α=cos2α-sin2α=2cos2α-1=1-2sin2α

2tanα
tan2α=—————
1-tan2α

sin3α=3sinα-4sin3α

cos3α=4cos3α-3cosα

3tanα-tan3α
tan3α=——————
1-3tan2α

三角函数的和差化积公式 三角函数的积化和差公式
α+β α-β
sinα+sinβ=2sin—--·cos—-—
2 2
α+β α-β
sinα-sinβ=2cos—--·sin—-—
2 2
α+β α-β
cosα+cosβ=2cos—--·cos—-—
2 2
α+β α-β
cosα-cosβ=-2sin—--·sin—-—
2 2 1
sinα ·cosβ=-[sin(α+β)+sin(α-β)]
2
1
cosα ·sinβ=-[sin(α+β)-sin(α-β)]
2
1
cosα ·cosβ=-[cos(α+β)+cos(α-β)]
2
1
sinα ·sinβ=- -[cos(α+β)-cos(α-β)]
2

化asinα ±bcosα为一个角的一个三角函数的形式(辅助角的三角函数的公式)

例题:
1、已知角α的终边在射线y=(-√3)x(x<0)上,求sinα+cosα的值。
1、射线y=(-√3)x(x<0)的斜率k=-√3=tanα ,
由公式得α=2π/3+2kπ,k∈N.
sinα=sin(2π/3+2kπ)=sin(2π/3)=√3/2.
cosα=cos(2π/3+2kπ)=cos(2π/3)=-1/2.
所以sinα+cosα=(√3-1)/2
2. 已知cosα=负三分之二,求:1+tan²α
cosα=负三分之二,(cosα)^2=4/9,得(sinα)^2=1-4/9=5/9

1+tan²α =1+(sinα/cosα)^2=1+sin²α/cos²α =1+(5/9)/(4/9)=9/4 已知cosα=负三分之二,求:1+tan²α

回答2:

太多了,不可能全给你发出来的!~

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