2a*cosθ+√3b*sinθ
∵√[(2a)^2+(√3b)^2]=√(4a^2+3b^2)
所以提取公因式√(4a^2+3b^2)得
√(4a^2+3b^2)*[2a/√(4a^2+3b^2) *cosθ+√3b/√(4a^2+3b^2)sinθ]
令2a/√(4a^2+3b^2)=sinA
则cosA=√(1-sin^2A)
=√[1-4a^2/√(4a^2+3b^2)^2]
=√[(4a^2+3b^2-4a^2)/√(4a^2+3b^2)^2]
=√3b/√(4a^2+3b^2)
∴原式化为√(4a^2+3b^2)*(sinAcosθ+cosAsinθ)
=√(4a^2+3b^2)*sin(A+θ)
∵sin(A+θ)≤1
∴
2a*cosθ+√3b*sinθ
=√(4a^2+3b^2)*sin(A+θ)
≤√(4a^2+3b^2)