3角函數 問題

Question

ｆ（θ）＝2（根號3）ｃｏｓ平方θ　+　ｓｉｎ2θ　

求ｍａｘ　　ｍｉｎ

　　　　

Answer 1

最佳解答

f(θ)=2√3cos^2(θ)+sin(2θ)

求max and min

f(θ)=2√3cos^2(θ)+sin(2θ)

=2√3cos^2(θ)-√3+sin(2θ)+√3

=√3(2cos^2(θ)-1)+sin(2θ)+√3

=√3cos(2θ)+sin(2θ)+√3

=2((√3/2)*cos(2θ)+(1/2)*sin(2θ))+√3

=2(sin(π/3)*cos2θ+cos(π/3)*sin2θ))+√3

=2(sin(π/3+2θ))+√3

當sin(π/3+2θ)=1, θ=π+π/12時有max

=2*1+√3

=2+√3

≒3.7320

當sin(π/3+2θ)=-1, θ=π+7π/12時有min

=-2+√3

≒-0.2679

答:max=2+√3, min=-2+√3

不懂請問!

2007-11-12 01:26:09 補充：

更正:

θ=nπ+π/12(n=N)

θ=nπ+7π/12(n=N)

加個n.