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题:设 f(x)=3sinx+4cosx,且-90°<α, β<90°,
f(α)=1,f(β)=2,求sin(α-β).
思路:显然,3sinα+4cosα=1,3sinβ+4cosβ=2,
∴ 4cosα=1-3sinα,4cosβ=2-3sinβ,
4sin(α-β)=sinα(2-3sinβ)-(1-3sinα)sinβ
=2sinα-sinβ.
又3sinα+4√[1-(sinα)^2]=1, 3sinβ+4√[1-(sinβ)^2]=2,
解得,50sinα=6±16√6,25sinβ=6±4√21,
∴ 4sin(α-β)=2sinα-sinβ=25(2sinα-sinβ)/25
=[±16√6-(±4√21)]/25,即sin(α-β)=[±4√6-(±√21)]/25.
又f(-π/2)=-3sinπ/2+4cosπ/2=-3,
f(0)=3sin0+4cos0=4,
f(x)=3sinx+4cosx=5cos(x-θ),其中sinθ=3/5,0<θ<π/4,
(当x=θ时,有f(x)max=5)
f(π/2)=3sinπ/2+4cosπ/2=3,∴α<β<0,sin(α-β)<0.
∴ sin(α-β)=(√21-4√6)/25. |
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