已知cos(π∕4+Θ)=-3∕5且π∕4

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已知cos(π∕4+Θ)=-3∕5且π∕4
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已知cos(π∕4+Θ)=-3∕5且π∕4
已知cos(π∕4+Θ)=-3∕5且π∕4

已知cos(π∕4+Θ)=-3∕5且π∕4
用a表示吧……
cos(π∕4+a)=-3∕5
sin[π/2-(π/4+a)]=sin(π/4-a)=-3/5
π∕4

∵π/4<θ<π/2,∴π/2<θ+π/4<3π/4
∵cos(π∕4+θ)=-3∕5
∴sin(π∕4+θ)=4∕5
∴cos(π∕4+θ)=sin(π/2-(π∕4+θ))=sin(π∕4-θ)=-3∕5
sin (π∕4+θ)=cos(π/2-(π∕4+θ))=cos(π∕4-θ)=4∕5
∴tan(π∕4-θ)=sin(π∕4-θ)/cos(π∕4-θ)=-3/4