
设α+β=2π/3,且0≤α≤π/2,求y=sinαsinβ的最大值和最小值
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y=sinαsinβ=1/2[cos(α-β)-cos(α+β)]
=1/2[cos(α-β)-cos(2π/3)]
=1/2[cos(α-β)+1/2]
=1/2cos(α-β)+1/4
又因为α+β=2π/3,且0≤α≤π/2,知-2π/3≤α-β≤π/2,
-1/2≤cos(α-β)≤1
0≤1/2cos(α-β)+1/4≤3/4
y=sinαsinβ的最大值是3/4\最小值是0.
=1/2[cos(α-β)-cos(2π/3)]
=1/2[cos(α-β)+1/2]
=1/2cos(α-β)+1/4
又因为α+β=2π/3,且0≤α≤π/2,知-2π/3≤α-β≤π/2,
-1/2≤cos(α-β)≤1
0≤1/2cos(α-β)+1/4≤3/4
y=sinαsinβ的最大值是3/4\最小值是0.
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