2a*cosθ+√3b*sinθ≤√(4a^2+3b^2) 怎么推出的
1个回答

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)