1.高分求各种三角恒等式

三角函数 sinx cosx tanx cotx secx cscx 含有与三角形三个内角有关的三角函数的恒等式,叫做三角恒等式 常见的三角恒等式及其证明 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