已知向量a=(sinθ,cosθ),b=(1,-2),且f(θ)=ab,求(2/3π)的值
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已知向量a=(sinθ,cosθ),b=(1,-2),且f(θ)=ab,求(2/3π)的值
已知向量a=(sinθ,cosθ),b=(1,-2),且f(θ)=ab,求(2/3π)的值
已知向量a=(sinθ,cosθ),b=(1,-2),且f(θ)=ab,求(2/3π)的值
f(θ)=a·b
=(sinθ,cosθ).(1,-2)
=sinθ-2cosθ
f(2π/3)= sin(2π/3)-2cos(2π/3)
=√3/2 + 1
输入法问题,用#表示那个字母了哈
f(#)=ab=1*sin#-2cos#
#=120°
所以原式=sin120°-2cos120°
=(根号3/2)-2*(-1/2)
=(根号3/2)+1
f(o)=ab=5^1/2xsin(120+arctan2)
=5^1/2sin(120+63.4)=-0.133