高中数学题目
已知f(θ)=(sinθ)^2+[sin(θ+α)]^2+[sin(θ+β)]^2,其中α、β为常数且0≤α≤β≤π,当且仅当α、β为何值时,f(θ)为与θ无关的定值.
解: 设f(θ)的定值为k,∵此时f(θ)与θ无关, 不妨分别取θ=0,π/2,得 f(0)=(sinα)^2+(sinβ)^2 ……(1) f(π/2)=1+(cosα)^2+(cosβ)^2 ……(2) 由(1)+(2),得 f(0)+f(π/2)=3 →2k=3,即k=3/2。
又,(2)-(1),得 1+cos2α+cos2β=0 ……(3) 由f(π/4)=1/2+[sin(π/4+α)]^2+[sin(π/4+β)]^2 =3/2+1/2×(sin2α+sin2β) =3/2 所以,sin2α+sin2β=0 ……(4) 解(3)、(4),得 cos2α=cos2β=-1/2 ∴α=π/3,β=2π/3,(0≤α≤β≤π) 将此结果代回原式,即有 f(θ)=(sinθ)^2+[sin(θ+π/3)]^2+[sin(θ+2π/3)]^2 =3/2×[(sinθ)^2+(cosθ)^2] =3/2 所以,当且仅当α=π/3,β=2π/3(0≤α≤β≤π)时, f(θ)为与θ无关的定值3/2。
我实在想不出更简洁的方法啊。
已知f(θ)=(sinθ)^2+[sin(θ+α)]^2+[sin(θ+β)]^2,其中α、β为常数且0≤α≤β≤π,当且仅当α、β为何值时,f(θ)为与θ无关的定值。
f(θ)=(sinθ)^2+[sin(θ+α)]^2+[sin(θ+β)]^2 =(sinθ)^2+(sinθcosα+cosθsinα)^2+(sinθcosβ+cosθsinβ)^2 =(sinθ)^2+sin^2 θcos^2 α+cos^2 θsin^2 α+2sinθcosθsinαcosα+sin^2 θcos^2 β+cos^2 θsin^2 β+2sinθcosθsinβcosβ =(sinθ)^2+sin^2 θcos^2 α+(1-sin^2 θ)sin^2 α+sin2θ*(1/2)sin2α+sin^2 θcos^2 β+(1-sin^2 θ)sin^2 β+sin2θ*(1/2)sin2β =(sin^2 θ)*(1+cos^2 α-sin^2 α+cos^2 β-sin^2 β)+sin^2 α+sin2θ*(1/2)sin2α+sin^2 β+sin2θ*(1/2)sin2β =(sin^2 θ)*(1+cos2α+cos2β)+(sinθcosθ)(sin2α+sin2β)+(sin^2 α+sin^2 β) 上式中,当1+cos2α+cos2β=0,且sin2α+sin2β=0时,f(x)与θ无关 ===> sin2α+sin2β=0,且cos2α+cos2β=-1 ===> (sin2α+sin2β)^2=0,且(cos2α+cos2β)^2=1 ===> sin^2(2α)+sin^2(2β)+2sin2αsin2β=0 且:cos^2(2α)+cos^2(2β)+2cos2αcos2β=1 两式相加得到: 1+1+2(sin2αsin2β+cos2αcos2β)=1 ===> sin2αsin2β+cos2αcosβ=-1/2 ===> cos(2α-2β)=-1/2 ===> 2α-2β=-2π/3,或者-4π/3 ===> α-β=-π/3,或者α-β=-2π/3……………………………………(1) 此时:f(θ)=sin^2 α+sin^2 β =[(1-cos2α)/2]+[(1-cos2β)/2] =1-(1/2)(cos2α+cos2β) =1-(1/2)*2cos(α+β)cos(α-β) =1-cos(α+β)cos(α-β) 这是一个定值,则它与α、β无关 所以,cos(α+β)=0【因为由前面(1)知,cos(a-β)≠0】 则,α+β=π/2,或者α+β=3π/2………………………………………(2) 由(1)(2)得到;α=π/12,β=5π/12。
答:详情>>