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a向量=(根号3cosx/2,2cosx/2),向量b=(2cosx/2,-sinx/2)函数f(x)=向量a·向量b
1)设θ∈[-π/2,π/2],且f(θ)=根号3+1,求θ的值
2)在△ABC中,AB=1,f(C)=根号3 +1,且△ABC的面积为 根号3/2,求sinA+sinB的值
人气:372 ℃ 时间:2019-08-19 04:32:44
解答
(1)
f(x) = a.b
=(√3cosx/2,2cosx/2).(2cosx/2,-sinx/2)
= 2√3(cosx/2)^2 - sinx
f(θ)=√3+1
√3+1 = 2√3(cosθ/2)^2 - sinθ
= √3( cosθ +1) - sinθ
1= √3cosθ - sinθ
1/2= (√3/2)cosθ - (1/2)sinθ
= sin(π/3+θ)
π/3+θ = π/6
θ = -π/6
(2) To be continued. .
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