求三个角以上的正、余弦和积互化公式

求三个角以上的正、余弦和积互化公式

sin(x+y+z)
=sinxcosycosz+cosxsinycosz+cosxcosysinz-sinxsinysinz
=cosxcosycosz(tanx+tany+tanz-tanxtanytanz)
cos(x+y+z)
=cosxcosycosz-cosxsinysinz-sinxcosysinz-sinxsinycosz
=cosxcosycosz(1-tanxtany-tanytanz-tanztanx)
tan(x+y+z)
=[tanx+tany+tanz-tanxtanytanz]/[1-tanxtany-tanytanz-tanztanx]
cot(x+y+z)
=[cotxcotycotz-cotx-coty-cotz]/[cotxcoty+cotycotz+cotzcotx-1]
已知sin(x+y+z)
=sinxcosycosz+cosxsinycosz+cosxcosysinz-sinxsinysinz
所以
sin(x+y-z)
=sinxcosycosz+cosxsinycosz-cosxcosysinz+sinxsinysinz
sin(x-y+z)
=sinxcosycosz-cosxsinycosz+cosxcosysinz+sinxsinysinz
sin(-x+y+z)
=-sinxcosycosz+cosxsinycosz+cosxcosysinz+sinxsinysinz
sin(x+y+z)
=sinxcosycosz+cosxsinycosz+cosxcosysinz-sinxsinysinz
所以4sinxsinysinz ＝sin(x+y-z) ＋sin(x-y+z) ＋sin(-x+y+z) －sin(x+y+z)
sin(x－y-z)
=sinxcosycosz－cosxsinycosz-cosxcosysinz－sinxsinysinz
sin(－x-y+z)
=－sinxcosycosz-cosxsinycosz+cosxcosysinz－sinxsinysinz
sin(-x+y+z)
=-sinxcosycosz+cosxsinycosz+cosxcosysinz+sinxsinysinz
sin(x+y+z)
=sinxcosycosz+cosxsinycosz+cosxcosysinz-sinxsinysinz
全相加
2cosxcosysinz－2sinxsinysinz ＝sin(x－y-z) ＋sin(－x-y+z) ＋sin(-x+y+z) ＋sin(x+y+z)
再利用4sinxsinysinz ＝sin(x+y-z) ＋sin(x-y+z) ＋sin(-x+y+z) －sin(x+y+z)
得到4cosxcosysinz＝ ＋sin(－x-y+z) ＋sin(-x+y+z) ＋sin(x+y-z) ＋sin(x-y+z)
其他自己仿照这2个,推理吧!

据我所知没有，我用的都是2个角三角函数公式，涉及多个角的都是2个角的变化形式。

sinθ+sinφ=2sin(θ/2+θ/2)cos(θ/2-φ/2)
sinθ-sinφ=2cos(θ/2+φ/2)sin(θ/2-φ/2)
cosθ+cosφ=2cos(θ/2+φ/2)cos(θ/2-φ/2)
cosθ-cosφ=-2sin(θ/2+φ/2)sin(θ/2-φ/2)
sinαsinβ=-1/2[cos(α+β)-cos(α-β)]

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sinθ+sinφ=2sin(θ/2+θ/2)cos(θ/2-φ/2)
sinθ-sinφ=2cos(θ/2+φ/2)sin(θ/2-φ/2)
cosθ+cosφ=2cos(θ/2+φ/2)cos(θ/2-φ/2)
cosθ-cosφ=-2sin(θ/2+φ/2)sin(θ/2-φ/2)
sinαsinβ=-1/2[cos(α+β)-cos(α-β)]
cosαcosβ= 1/2[cos(α+β)+cos(α-β)]
sinαcosβ= 1/2[sin(α+β)+sin(α-β)]
cosαsinβ= 1/2[sin(α+β)-sin(α-β)]

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sinθ+sinφ=2sin(θ/2+θ/2)cos(θ/2-φ/2)
sinθ-sinφ=2cos(θ/2+φ/2)sin(θ/2-φ/2)
cosθ+cosφ=2cos(θ/2+φ/2)cos(θ/2-φ/2)
cosθ-cosφ=-2sin(θ/2+φ/2)sin(θ/2-φ/2)
sinαsinβ=-1/2[cos(α+β)-cos(α-β)]

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sinθ+sinφ=2sin(θ/2+θ/2)cos(θ/2-φ/2)
sinθ-sinφ=2cos(θ/2+φ/2)sin(θ/2-φ/2)
cosθ+cosφ=2cos(θ/2+φ/2)cos(θ/2-φ/2)
cosθ-cosφ=-2sin(θ/2+φ/2)sin(θ/2-φ/2)
sinαsinβ=-1/2[cos(α+β)-cos(α-β)]
cosαcosβ= 1/2[cos(α+β)+cos(α-β)]
sinαcosβ= 1/2[sin(α+β)+sin(α-β)]
cosαsinβ= 1/2[sin(α+β)-sin(α-β)]

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