若sin(3派/4+α)=5/13,cos(派/4-α)=3/5,0
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若sin(3派/4+α)=5/13,cos(派/4-α)=3/5,0
若sin(3派/4+α)=5/13,cos(派/4-α)=3/5,0
若sin(3派/4+α)=5/13,cos(派/4-α)=3/5,0
0
哪里有β?
应该是cos(π/4-β)=3/5
3π/4<3π/4+α<π
所以cos(3π/4+α)<0
sin²(3π/4+α)+cos²(3π/4+α)=1
所以cos(3π/4+α)=-12/13
π/4<β<3π/4
-π/2<π/4-β<0
所以sin(π/4-β)<0
sin²(π/...
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哪里有β?
应该是cos(π/4-β)=3/5
3π/4<3π/4+α<π
所以cos(3π/4+α)<0
sin²(3π/4+α)+cos²(3π/4+α)=1
所以cos(3π/4+α)=-12/13
π/4<β<3π/4
-π/2<π/4-β<0
所以sin(π/4-β)<0
sin²(π/4-β)+cos²(π/4-β)=1
sin(π/4-β)=-4/5
cos(α+β)
=sin[π/2-(α+β)]
=sin[(3π/4+α)-(π/4-β)]
=sin(3π/4+α)cos(π/4-β)-cos(3π/4+α)sin(π/4-β)
=-33/65
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