Question

三角函数 已知sinA+sinB+sinC=0 cosA+cosB+cosC=0 求证：sin2A+sin2B+sin2C=0,COS2A+COS2B+COS2C=0

在帮忙做一道，证明sinA+1/2sin2A+1/3sin3A=0

Answer 1

1.(sinA)^2+(cosA)^2=(sinB+sinC)^2+(cosB+cosC)^2=1
得2cos(B-C)=-1
得cos(B-C)=-1/2
sin2A+sin2B+sin2C=2(sinB+sinC)(cosB+cosC)+sin2B+sin2C=2[sin2B+sin2C+sin(B+C)]=2sin(B+C)[2cos(B-C)+1]=0
2.cos2A+cos2B+cos2C=2(cosA)^2-1+cos2B+cos2C=2(cosB)^2+2(cosC)^2-1+4cosBcosC+cos2B+cos2C=2cos2B+2cos2C+4cosBcosC+1=4cos(B+C)cos(B-C)+2[cos(B+C)+cos(B-C)]+1=-2cos(B+C)+2cos(B+C)-1+1=0
3.sinA+1/2sin2A+1/3sin3A=0显然是错题...
当A=π/2时,sinA=1,sin2A=0,sin3A=-1.
所以sinA+1/2sin2A+1/3sin3A=1-1/3=2/3
所以是错题目

Answer 2

上述两式相乘有
SINA*COSA+SINB*COSB+SINC*COSC+(SINA*COSB+COSA*SINB)+(SINB*COSC+COSB*SINC)+(SINC*COSA+COSC*SINA)
=(SIN2A+SIN2B+SIN2C)/2+SIN(A+B)+SIN(B+C)+SIN(A+C)=0
这里SIN(A+B)+SIN(B+C)+SIN(A+C)
令A+B+C=U;则该式子有
SIN(U-A)+SIN(U-B)+SIN(U-C)
=SINUCOSA-COSUSINA+SINUCOSB-COSUSINB+SINUCOSC-COSUSINC
=SINU(COSA+COSB+COSC)-COSU(SINA+SINB+SINC)=0
于是
SIN2A+SIN2B+SIN2C=0,

题干中给的两个式子分别两边平方：
SINA^2+SINB^2+SINC^2+2*(SINA*SINB+SINA*SINC+SINC*SINB)=0
COSA^2+COSB^2+COSC^2+2*(COSA*COSB+COSA*COSC+COSC*COSB)=0
之后下面的式子减去上面的，经过化简，可得：
(COS2A+COS2B+COS2C)/2+COS(A+B)+COS(B+C)+COS(A+C)=0
还是用U
COS(A+B)+COS(B+C)+COS(A+C)
=COS(U-C)+COS(U-A)+COS(U-B)
=COSU*COSC+SINU*SINC+COSU*COSA+SINU*SINA+COSU*COSB+SINU*SINB
=COSU(COSA+COSB+COSC)+SINU(SINA+SINB+SINC)=0
于是
COS2A+COS2B+COS2C=0

Answer 3

sinA+sinB=-sinC cosA+cosB=-cosC 再平方
相减 就 出现cos2c
依次下去
就能得出了！