Question

高中数学（三角函数），求解答，有详细的过程

Answer 1

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追答

Answer 2

以O为原点,射线OP为x轴正半轴建系,则
C(cosα,sinα),OQ:y=√3x
BC=sinα,D(sinα/√3,sinα)
∴CD=cosα-sinα/√3
S=BC*CD=sinαcosα-sin²α/√3
=1/2*[sin2α-(1-cos2α)/√3]
=1/2*(sin2α+1/√3*cos2α-1/√3)
=1/2*[2/√3*sin(2α+π/6)-1/√3]
∵α∈(0,π/3),∴2α+π/6∈(π/6,5π/6)
∴当α=π/6时,sin(2α+π/6)有最大值1
∴Smax=1/2*(2/√3-1/√3)=1/2√3=√3/6

Answer 3

BC = OC * sinα = sinα = AD
AD / AO = sin60° = √3
AO = sinα / √3
BO = OC * cosα = cosα
AB = BO - AO = cosα - sinα / √3
S□ABCD = AB * BC = sinα * cosα - (sinα)^2 / √3
= 1/2 * sin2α - (1 - cos2α) / 2√3
= (1/√3) * [ √3/2 * sin2α + 1/2 * cos2α] - 1 / 2√3
= (1/√3) * sin(2α + 30°) - 1 / 2√3
当α=30°时上式取得最大值1/2√3