已知向量a=[cos(3x/2),sin(3x/2)],b=[cos(x/2),-sin(x/2),]且x∈[0,π/2]
(1)求a·b及|a+b|;
(2)求函数f(x)=a·b-|a+b|的最小值及此时x的值
人气:312 ℃ 时间:2019-12-13 00:12:14
解答
a·b=cos(3x/2)*cos(x/2)-sin(3x/2)*)sin(x/2)=cos(3x/2+x/2)=cos2x,向量a+b=[cos(3x/2)+cos(x/2),sin(3x/2)-sin(x/2)]=(2cosxcosx/2,2cosxsinx/2),|a+b|=√[(2cosxcosx/2)^2+(2cosxsinx/2)^2]=2√{(cosx)^2[(sinx/2...
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