神马作文网已知sina+3cosa=2,则（sina-cosa)\(sina+cosa)=
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已知sina+3cosa=2,则（sina-cosa)\(sina+cosa)=
sinA+3cosA=2
--->2sin(A/2)cos(A/2)+3[cos(A/2)^2-3(sin(A/2)]^2=2[cos(A/20]^2+[sin(A/2)]^2
--->4[sin(A/2)]^2+2sin(A/2)cos(A/)-2[cos(A/2)]^2=0【两边同除2[cos(A/2)]^2】
--->2[tan(A/2)]^2+tan(A/2)-1=0
--->tan(A/2)=-1或者-1/2.
tan(A/2)=-1时,sinA=2tan(A/2)/{1+[tan(A/2]^2}=-1,cosA=0
--->(sinA-cosA)/(sinA+cosA)=(-1-0)/(-1+0)=1
tan(A/2)=1/2时,tanA=2tan(A/2)/{1-[tan(A/2)]^2}=4/3
--->(sinA-cosA)/(sinA+cosA)
=(tanA-1)(tanA+1)
=(4/3-1)/(4/3+1)
=1/7.
--->2sin(A/2)cos(A/2)+3[cos(A/2)^2-3(sin(A/2)]^2=2[cos(A/20]^2+[sin(A/2)]^2
--->4[sin(A/2)]^2+2sin(A/2)cos(A/)-2[cos(A/2)]^2=0【两边同除2[cos(A/2)]^2】
--->2[tan(A/2)]^2+tan(A/2)-1=0
--->tan(A/2)=-1或者-1/2.
tan(A/2)=-1时,sinA=2tan(A/2)/{1+[tan(A/2]^2}=-1,cosA=0
--->(sinA-cosA)/(sinA+cosA)=(-1-0)/(-1+0)=1
tan(A/2)=1/2时,tanA=2tan(A/2)/{1-[tan(A/2)]^2}=4/3
--->(sinA-cosA)/(sinA+cosA)
=(tanA-1)(tanA+1)
=(4/3-1)/(4/3+1)
=1/7.
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