已知向量a=(cos3/2x,sin3/2x),b向量=(cosx/2,-sinx/2),且x属于[0，π/2]

显示全部楼层 · 2012-4-22 16:51:06

解：1. 以下略去“向量”二字。 a.b=cos(3x/2)cos(x/2)-sin(3x/2)(sin(x/2).
=cos(3x/2+x/2).
=cos2x.
x∈[0, π/2],
2x∈[0,π] |a+b|^2=(a+b)^2.
=a^2+2ab+b^2.
=1+2|a||b|cos+1.
=2+2cos,a,b>
|a|=1. |b|=1.
=2(1+cos).
=2*2cos^2[()/2]|a+b|=2cos(<...

千问 · 2012-4-22 16:51:06

(1) a?b = cos(3/2)xcos(x/2) - sin(3/2)xsin(x/2) = cos2x 　 |a+b| = √(a2+b2+2a?b) = √(2+2cos2x) = 2cosx (2) f(x)=cos2x-4mcosx=2cos2x-4mcosx-1=2(cosx-m)...