Question

已知SINA+SINB+SINC=0,COSA+COSB+COSC=0,求证SIN2A+SIN2B+SIN2C=0,COS2A+COS2B+COS2C=0

要详细的过程啊

Answer 1

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)=0COSA^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UCOS(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