三角恒等式

设A，B，C是三角形的三个内角

(1)

tanA＋tanB＋tanC=tanAtanBtanC

证明：

tanA+tanB+tanC=tan(A+B)(1-tanAtanB)+tanC=tan(π-c)(1-tanAtanB)+tanC=-tanC(1-tanAtanB)+tanC=tanAtanBtanC

(2)

cotAcotB＋cotBcotC+cotCcotA=1

证明：

tanA＋tanB＋tanC=tanAtanBtanC

cotX*tanX=1

tanA*cotAcotBcotC＋tanB*cotAcotBcotC＋tanC*cotAcotBcotC=tanAtanBtanC*cotAcotBcotC

cotAcotB＋cotBcotC+cotCcotA=1

(3)

(cosA)^2+(cosB)^2+(cosC)^2+2cosAcosBcosC=1

证明：

(cosA)^2+(cosB)^2+x^2+2cosAcosBx=1

x^2+2cosAcosBx+(cosA)^2+(cosB)^2-1=0

x={-2cosAcosB+-√[(2cosAcosB)^2-4((cosA)^2+(cosB)^2-1)]}/2

x=-cosAcosB+-√[(cosAcosB)^2-((cosA)^2+(cosB)^2-1)]

x=-cosAcosB+-√[1-(cosA)^2][1-(cosB)^2]

x=-cosAcosB+-√[(sinA)^2(sinB)^2]

x=-cosAcosB+-sinAsinB

x=-cos(A+B)或x=-cos(A-B)

x=cosC或x=-cos(A-B)

所以

cosC是方程的一个根

所以

(cosA)^2+(cosB)^2+(cosC)^2+2cosAcosBcosC=1

(4)

cosA+cosB+cosC＝1＋4sin(A/2)sin(B/2)sin(C/2)

证明：

cosA+cosB+cosC=1+4sin(A/2)sin(B/2)sin(C/2)

cos(180-B-C)+cosB+cosC=1+2sin(A/2)[2sin(B/2)sin(C/2)]

cos(180-B-C)+cosB+cosC=1+2cos(B/2+C/2)[2sin(B/2)sin(C/2)]

-cos(B+C)+cosB+cosC=1+2cos(B/2+C/2)[2sin(B/2)sin(C/2)]

-cos(B+C)+cosB+cosC=1+2cos(B/2+C/2)[cos(B/2-C/2)-cos(B/2+C/2)]

-cos(B+C)+cosB+cosC=1+2cos(B/2+C/2)cos(B/2-C/2)-2[cos(B/2+C/2)]^2

cosB+cosC=2cos(B/2+C/2)cos(B/2-C/2)

2[cos(B/2+C/2)]^2-1=cos(B+C)

(5)

tan(A/2)tan(B/2)+tan(B/2)tan(C/2)+tan(C/2)tan(A/2)=1

证明：

A/2+B/2+C/2=π/2

(π/2-A)+(π/2-B)+(π/2-C)=π

cot(π/2-A)cot(π/2-B)+cot(π/2-C)cot(π/2-B)+cot(π/2-A)cot(π/2-C)=1

tan(A/2)tan(B/2)+tan(B/2)tan(C/2)+tan(C/2)tan(A/2)=1

(6)

sin2A+sin2B+sin2C=4sinAsinBsinC

证明：

设三角形ABC的外心为O

S△ABO+S△ACO+S△CBO=S△ABC

(1/2)RRsin2C+(1/2)RRsin2B+(1/2)RRsin2A=(1/2)bcsinA=(1/2)2RsinB*2RsinC*sinA

sin2A+sin2B+sin2C=4sinAsinBsinC

(7)

sinA+sinB+sinC=4cos(A/2)cos(B/2)cos(C/2)

证明：

4cos(A/2)cos(B/2)cos(C/2)

=[2cos(C/2)]*[2cos(A/2)cos(B/2)]

=[2sin(A/2+B/2)]*[cos(A/2+B/2)+cos(A/2-B/2)]

=2sin(A/2+B/2)cos(A/2+B/2)+2sin(A/2+B/2)cos(A/2-B/2)

=sin(A+B)+2sin(A/2+B/2)cos(A/2-B/2)

=sinC+2sin[(A+B)/2]cos[(A-B)/2]

=sinC+sin[(A+B)/2+(A-B)/2]+sin[(A+B)/2-(A-B)/2]

=sinC+sinA+sinB