已知向量a=(cos 5x/3,sin 5x/3),向量b=(cos x/3,—sin x/3),x∈[0,π/2]
问题描述：
已知向量a=(cos 5x/3,sin 5x/3),向量b=(cos x/3,—sin x/3),x∈[0,π/2]
求|a+b|
求a×b
答
向量a=(cos5x/3,sin5x/3),向量b=(cosx/3,-sinx/3)
则：a+b=(cos5x/3+cosx/3,sin5x/3-sinx/3)
|a+b|^2=(cos5x/3+cosx/3)^2+(sin5x/3-sinx/3)^2=2+2(cos5x/3cosx/3-sin5x/3sinx/3)
=2+2cos2x=4cosx^2,因为：x∈[0,π/2],所以cosx>0,故：|a+b|=2cosx
求a×b是求数量积吧：
a dot b=(cos5x/3,sin5x/3)dot(cosx/3,-sinx/3)
=cos5x/3cosx/3-sin5x/3sinx/3=cos2x为什么2+2cos2x=4cosx^2是哈，2+2cos2x=2+2(2cosx^2-1)=2+4cosx^2-2=4cosx^2------------三角函数的倍角公式，看来你需要努力了。cos2x=2cosx^2-1=1-2sinx^2=cosx^2-sinx^2