已知2cosx+3sinx=(根号26)/2,求tan2x
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已知2cosx+3sinx=(根号26)/2,求tan2x
![已知2cosx+3sinx=(根号26)/2,求tan2x](/uploads/image/z/3773775-39-5.jpg?t=%E5%B7%B2%E7%9F%A52cosx%2B3sinx%3D%28%E6%A0%B9%E5%8F%B726%29%2F2%2C%E6%B1%82tan2x)
2cosx+3sinx=(√26)/2,
(2/√13)cosx+(3/√13)sinx=(√2)/2.
设2/√13=siny,3/√13=cosy,则
sinycosx+cosysinx=(√2)/2,sin(x+y)=(√2)/2,x+y=π/4+2kπ,x=π/4-y+2kπ,
tanx=tan(π/4-y)=[tan(π/4)-tany]/[1+tan(π/4)tany]=(1-2/3)/(1+2/3)=1/5.
∴tan2x=2tanx/[1-(tanx)^2]=(2/5)/[1-(1/5)^2]=5/12.
(2/√13)cosx+(3/√13)sinx=(√2)/2.
设2/√13=siny,3/√13=cosy,则
sinycosx+cosysinx=(√2)/2,sin(x+y)=(√2)/2,x+y=π/4+2kπ,x=π/4-y+2kπ,
tanx=tan(π/4-y)=[tan(π/4)-tany]/[1+tan(π/4)tany]=(1-2/3)/(1+2/3)=1/5.
∴tan2x=2tanx/[1-(tanx)^2]=(2/5)/[1-(1/5)^2]=5/12.
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