Question

sina+sinb+sinc=0, cosa+cosb+cosc=0,则cos(a-b)的值为?

sina+sinb+sinc=0, cosa+cosb+cosc=0,则cos(a-b)的值为?麻烦写下详细过程.并提点思维方法...

Answer 1

cos(a-b)=cosacosb+sinasinb

因为sina+sinb+sinc=0
移项得sina+sinb=-sinc
两边都平方得1+2sinasinb=sin^2c

同理1+2cosacosb=cos^2c

所以
cos(a-b)
=(sin^2c-1+cos^2c-1)/2
=[(sin^2c+cos^2c)-2]/2
=-1/2

Answer 2

因为(sinc)^2=(sina+sinb)^2=(sina)^2+2sina*sinb+(sinb)^2,
同理(cosc)^2=(cosa)^2+2cosa*cosb+(cosb)^2,
所以相加得1=1+2(cosa*cosb+sina*sinb)+1,
所以cos(a-b)=-1/2。

Answer 3

sina+sinb=-sinc
cosa+cosb=-cosc 分别平方相加 再展开cos(a-b),能得出值