设cos (x +π/4 )=3/5,17π/12
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设cos(x+π/4)=3/5,17π/12设cos(x+π/4)=3/5,17π/12设cos(x+π/4)=3/5,17π/12∵17π/12sinx+cosx=-4√2/5.(2)∴由(2)-(
设cos (x +π/4 )=3/5,17π/12
设cos (x +π/4 )=3/5,17π/12
设cos (x +π/4 )=3/5,17π/12
∵17π/12sinx+cosx=-4√2/5.(2)
∴由(2)-(1)得2sinx=-7√2/5 ==>sinx=-7√2/10
由(2)+(1)得2cosx=-√2/5 ==>cosx=-√2/10
故(sin2x+2sin²x)/(1-tan x)=(2sinxcosx+2sin²x)/(1-sinx/cosx)
=2sinx(cosx+sinx)/(1-sinx/cosx)
=2(-7√2/10)(-√2/10-7√2/10)/(1-(-7√2/10)/(-√2/10))
=-28/75
设cos (x +π/4 )=3/5,17π/12
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