sin,cos與sec,csc有什么等式關(guān)系


推選答案sin²a+cos²a=1
1+tan²a=sec²a
1+cot²a=csc²a
sina/cosa=tana
seca/csca=tana
cosa/sina=cota
sin2α = 2cosαsinα
= sin²(α+π/4)-cos²(α+π/4)
= 2sin²(a+π/4)-1
= 1-2cos²(α+π/4)
cos2α = cos²α-sin²α
= 1-2sin²α
= 2cos²α-1
= 2sin(α+π/4)·cos(α+π/4)
tan2α = 2tanα/[1-(tanα)²]
sin3α = 3sinα-4sin³α
cos3α = 4cos³α-3cosα
tan3α = (3tanα-tan³α)/(1-3tan²α)
sin3α = 4sinα·sin(π/3-α)·sin(π/3+α)
cos3α = 4cosα·cos(π/3-α)·cos(π/3+α)
tan3α = tanα·tan(π/3-α)·tan(π/3+α)
sin(nα) = ncos(n-1)α·sinα - C(n,3)cos(n-3)α·sin3α + C(n,5)cos(n-5)α·sin5α-…
cos(nα) = cosnα - C(n,2)cos(n-2)α·sin2α + C(n,4)cos(n-4)α·sin4α
sin,cos與sec,csc有什么等式關(guān)系

Asinα+Bcosα = √(A2+B2)sin[α+arctan(B/A)]
Asinα+Bcosα = √(A2+B2)cos[α-arctan(A/B)]
sin(α/2) = ±√[(1-cosα)/2]
cos(α/2) = ±√[(1+cosα)/2]
tan(α/2) = ±√[(1-cosα)/(1+cosα)]=sinα/(1+cosα)=(1-cosα)/sinα=cscα-cotα
cot(α/2) = ±√[(1+cosα)/(1-cosα)]=(1+cosα)/sinα=sinα/(1-cosα)=cscα+cotα
sec(α/2) = ±√[(2secα/(secα+1)]
csc(α/2) = ±√[(2secα/(secα-1)]
kπ+a
sin(2kπ+α)=sinα
cos(2kπ+α)=cosα
tan(kπ+α)=tanα
cot(kπ+α)=cotα
sec(2kπ+α)=secα
csc(2kπ+α)=cscα
sin(π+α)=-sinα
cos(π+α)=-cosα
tan(π+α)=tanα
cot(π+α)=cotα
sec(π+α)=-secα
csc(π+α)=-cscα

-a
sin(-α)=-sinα
cos(-α)=cosα
tan(-α)=-tanα
cot(-α)=-cotα
sec(-α)=secα
csc(-α)=-cscα

π-a
sin(π-α)=sinα
cos(π-α)=-cosα
tan(π-α)=-tanα
sin,cos與sec,csc有什么等式關(guān)系

cot(π-α)=-cotα
sec(π-α)=-secα
csc(π-α)=cscα

π/2±a
sin(π/2+α)=cosα
cos(π/2+α)=-sinα
tan(π/2+α)=-cotα
cot(π/2+α)=-tanα
sec(π/2+α)=-cscα
csc(π/2+α)=secα
sin(π/2-α)=cosα
cos(π/2-α)=sinα
tan(π/2-α)=cotα
cot(π/2-α)=tanα
sec(π/2-α)=cscα
csc(π/2-α)=secα

3π/2±a
sin(3π/2+α)=-cosα
cos(3π/2+α)=sinα
tan(3π/2+α)=-cotα
cot(3π/2+α)=-tanα
sec(3π/2+α)=cscα
csc(3π/2+α)=-secα
sin(3π/2-α)=-cosα
cos(3π/2-α)=-sinα
tan(3π/2-α)=cotα
cot(3π/2-α)=tanα
sec(3π/2-α)=-cscα
csc(3π/2-α)=-secα
tan(a+π/4)=(tan a+1)/(1-tan a)
tan(a-π/4)=(tan a-1)/(1+tan a)
asinx+bcosx=[√(a²+b²)]{[a/√(a²+b²)]sinx+[b/√(a²+b²)]cosx}=[√(a²+b²)]sin(x+y)
sin,cos與sec,csc有什么等式關(guān)系

sinα·cosβ=(1/2)[sin(α+β)+sin(α-β)]
cosα·sinβ=(1/2)[sin(α+β)-sin(α-β)]
cosα·cosβ=(1/2)[cos(α+β)+cos(α-β)]
sinα·sinβ= -(1/2)[cos(α+β)-cos(α-β)]
sinA+sinB+sinC=4cos(A/2)cos(B/2)cos(C/2)
cosA+cosB+cosC=1+4sin(A/2)sin(B/2)sin(C/2)
tanA+tanB+tanC=tanAtanBtanC
cot(A/2)+cot(B/2)+cot(C/2)=cot(A/2)cot(B/2)cot(C/2)
tan(A/2)tan(B/2)+tan(B/2)tan(C/2)+tan(C/2)tan(A/2)=1
cotAcotB+cotBcotC+cotCcotA=1
(cosA)^2+(cosB)^2+(cosC)^2+2cosAcosBcosC=1
sin2A+sin2B+sin2C=4sinAsinBsinC
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