cos(α+β)=cosαcosβ-sinαsinβ
cos(α-β)=cosαcosβ+sinαsinβ
sin(α+β)=sinαcosβ+cosαsinβ
sin(α-β)=sinαcosβ-cosαsinβ
tan(α+β)=(tanα+tanβ)/( 1-tanαtanβ)
tan(α-β)=(tanα-tanβ)/( 1+tanαtanβ)
sin2α
=
2sinαcosα
cos2α
=
(cosα)^2
-
(sinα)^2=2(cosα)^2
- 1= 1-2(sinα)^2
tan2α=2tanα/[ 1-(tanα)^2]
2, sine angle formula:
sin2α
=
2cosαsinα
Deduction: SIN2A = SIN (a+a) = SINA COSA+COSA SINA = 2 SINA COSA.
Extended formula: sin2a = 2sinacosa = 2tanacosa2 = 2tana/[1+tana2]
Cosine double angle formula:
The cosine double angle formula has three sets of expressions, which are equivalent:
1.cos2a=cosa^2-sina^2=[ 1-tana^2]/[ 1+tana^2]
2.Cos2a= 1-2Sina^2
3.Cos2a=2Cosa^2- 1
Deduction: cos2a = cos (a+a) = cosacosa-sinasina = (cosa) 2-(sina) 2 = 2 (cosa) 2-1
= 1-2(sinA)^2
Tangent dihedral formula:
tan2α=2tanα/[ 1-(tanα)^2]
Deduction: tan2a = tan (a+a) = (tana+tana)/(1-tanatana) = 2tana/[1-(tana) 2]
3. Auxiliary angle formula:
For the function of type acosx+bsinx, we can deform Acosx+BSINX = SQRT (A2+B2) (Acosx/SQRT (A2+B2)+BSINX/SQRT (A2+B2)) so that the point (b, a) is definite.
∴acosx+bsinx=sqrt(a^2+b^2)sin(x+arctan(a/b))
This is the auxiliary angle formula.
Let the formula to be proved be acosa+bsina = √ (a 2+b 2) sin (a+m).
(tanM=a/b)
The following is the proof process:
Let acosA+bsinA=xsin(A+M)
∴acosa+bsina=x((a/x)cosa+(b/x)sina)
Problem, (a/x) 2+(b/x) 2 = 1, SINM = a/x, COSM = B/X.
∴x=√(a^2+b^2)
∴acosA+bsinA=√(a^2+b^2)sin(A+M)
，tanM=sinM/cosM=a/b