设矩阵 sin2a sina+cosa
问题描述:
设矩阵 sin2a sina+cosa
设矩阵 sin2a sina+cosa a 1/2
( ) = ( )
cos2a sina-cosa b c
且0
答
sina+cosa =0.5,a=sin2a=(sina+cosa)^2-1=0.25-1=-0.75
b=cos2a=cos^2a-sin^2a=(cosa+sina)(cosa-sina)=0.5(cosa-sina)=-0.5c
sin2a〈0,a在第二区间,sina-cosa〉0
c^2=(sina-cosa)^2=1-sin2a=1-a=1.75,
写结果