matlab y=cos(m*cos-1(x))答案

A+B+C=πA+B=π-CM⊥N=>M.N=0(2cosC/2,-sin(A+B)).(cosC/2,2sin(A+B))=0(cosC/2)^2-(sin(A+B))^2=0(cosC/2)^2-

m.n=sinA-2COS^2(B+C）=sinA-cos（π-B-C）²=sinA-cos²A=sin²A+sinA-1=（sinA+1/2）²-5/4当si

∵m//n∴﹙√3b-c﹚×cosA－a×cosC＝0,根据余弦定理得,﹙√3b-c﹚×﹙b²＋c²－a²﹚／2bc＝a×﹙a²＋b²－c²

向量m*n=1/2-cosA/2*cosA/2+sinA/2sinA/2=1/2cos^2(A/2)-sin^2(A/2)=-1/2cosA=-1/2A=120度S=1/2bcsinA=√3bc*√3

cos(a-b)cos(b-c)+sin(a-b)sin(b-c)=cos[(a-b)-(b-c)]=cos(a+c-2b)

1.mn=1/2[cos(A/2),sin(A/2)]*[-cos(B/2),sin(B/2)]=-cos(A/2)cos(B/2)+sin(A/2)sin(B/2)=-cosC=60度2.根号3)/

因为向量m⊥向量n所以,2cosc/2*cosc/2-sinc*2sinc=0可得,2(cosc/2)^2-2(sinc)^2=01+cosc-2*(1-cosc^2)=0cosc=1/2c=π/3(

/>∵m//n∴[2cos(B-C)-1]/cosBcosC=4/12cos(B-C)-1=4cosBcosC2(cosBcosC+sinBsinC)-1=4cosBcosC2cosBcosC+2si

cos(θ/2+π/8)=-4/5m+n=(cosθ+√2-sinθ,sinθ+cosθ)丨m+n丨=√[(cosθ+√2-sinθ)^2+(sinθ+cosθ)^2]=√[4+2√2cosθ-2√2

SIN(B+C）=sinBcosC+cosBsinCCOS(B+C)=cosBcosC-sinBsinC

cosC/cosB=（3a-c）/b用余弦定理：【（a^2+b^2-c^2）/2ab】/【（a^2+c^2-b^2）/2ac】=（3a-c）/b化简后得：2ac=3a^2+3c^2-3b^2（a^2+

∵m∥n∴满足m=入n(入≠0)∴sinB=入cos2B,√3=入[4cos²(B/2)-2]∴cos2B/sinB=2*[2cos²(B/2)-1]/√3即2sinB*[2cos

sin²α+cos²α=1sinα=±√(1-m²)tanα=sinα/conα=±√(1-m²)/m

cos²B-cos²C=sin²Acos²B=1-sin²Bcos²C=1-sin²Csin²C-sin²B=

m垂直n=>m.n=0(2cos(C/2),-sinC).(cos(C/2),2sinC)=02(cos(C/2))^2-2(sinC)^2=0cosC+1-2(sinC)^2=0cosC+1-2(1

你的式子有一项好像抄错了如果原题是求证a²(cos2B-cos2C)+b²(cos2C-cos2A)+c²(cos2A-cos2B)=0的话证明如下：a²(co

sin(120-a)=sin(90+(30+a))=-cos(30-a)=-m

向量m+向量n=(cosB+√2-sinB,sinB+cosB).|向量m+向量n|=（8根号2）/5,则有(cosB+√2-sinB)^2+(sinB+cosB)^2=[(8根号2）/5]^2,√2

A=B=C是等边三角形

mn=sin(A-B)+2cosAsinB=sinAcosB+cosAsinB=sin(A+B)=sinC=-sin2C=-2sinCcosC,∴cosC=-1/2,C=120°,A+B=60°,si