已知π/4

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已知π/4已知π/4已知π/4∵π/4<α<3π/4,∴π/2<π/4+α<π,而cos(π/4+α)=-3/5,∴sin(π/4+α)=√[1-(-3/5)^2]=4/5.∵0<β<π/4,∴3π/

已知π/4
已知π/4

已知π/4
∵π/4<α<3π/4,∴π/2<π/4+α<π,而cos(π/4+α)=-3/5,
∴sin(π/4+α)=√[1-(-3/5)^2]=4/5.
∵0<β<π/4,∴3π/4<3π/4+β<π,而sin(3π/4+β)=5/13,
∴cos(3π/4+β)=-√[1-(5/13)^2]=-12/13.
∴sin(α+β)=-sin[π+(α+β)]=-sin[(π/4+α)+(3π/4+β)]
=-sin(π/4+α)cos(3π/4+β)-cos(π/4+α)sin(3π/4+β)
=-(4/5)(-12/13)-(-3/5)(5/13)=(48+15)/(5×13)=63/(5×13)
=63/65.