cos(x-π/4)=cos[(x+π/4)-π/2]=sin(x+π/4)=-√[1-cos²(x+π/4)]=-√[1-(3/5)²]=-4/5
sin(2x)=-cos(2x+π/2)=-cos[2(x+π/4)]=1-2cos²(x+π/4)=1-2•(3/5)²=7/25
[sin(2x)+2sin²x]/(1-tanx)
=2(sinxcosx+sin²x)/(1-sinx/cosx)
=2(cosx+sinx)/(1/sinx-1/cosx)
=2(cosx+sinx)sinxcosx/(cosx-sinx)
=cos(x-π/4)sin(2x)/cos(x+π/4)
=-4/5•7/25/(3/5)
=-28/75追问
是7π，怎么17π也对
追答因7π/12＜x＜7π/4,所以5π/6＜π/4+x＜2π
因为cos(π/4+x)＝3/5 ＞0
所以tan(π/4+x)＝－4/5
sin2x＝－cos(2x＋π/2)＝1－2[cos(π/4＋x)]^2＝7/25
原式＝sin2x(1＋tanx)/(1-tanx)＝sin2x×tan(π/4+x）
＝(7/25)×(-4/5)＝-28/125
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已知cos(π\/4 +x)=3\/5,17π\/12<x<7π\/4,求(sin2x+2sin�0�5x)\/...
您好：解答如下17π\/12<x<7π\/4，所以5π\/3<π\/4 +x<2π得到cos(π\/4 +x)=3\/5，sin(π\/4 +x)=-4\/5cosx=cos(π\/4 +x-π\/4 ) =cos(π\/4 +x)cosπ\/4+sin(π\/4 +x)sinπ\/4 =3\/5×√2\/2+（-4\/5）×√2\/2 =-√2\/10因为17π\/12<x<7π\/4，所以sin...
若cos(π\/4+x)=3\/5,17π\/12<x<7π\/4,求sin2x+2cos²x\/1-tanx
5π\/3<π\/4+x<2π,由此可见π\/4+x属于第四象限。因此有sin(π\/4+x)=-4\/5 将相应地cos(π\/4+x)=3\/5展开可得：cosx-sinx=根号2*3\/5 将相应地sin(π\/4+x)=-4\/5展开可得：cosx+sinx=-根号2*4\/5 从而可得cosx=-根号2\/10，sinx=-根号2*7\/10，tanx=sinx\/cosx=7\/2 相应地...
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cos(π\/4+x)=3\/5,17π\/12<x<7π\/4,求sin2x+2sin^2x\/1-tan^2x
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cos(π\/4+X)=3\/5,17π\/12<X<7π\/4,[sin2X+2(sinX)^2]\/(1-tanX)
COSX-SinX=3\/5*2^0.5;cosX*sinX=7\/50;CosX+sinX=-4\/5*2^0.5,[sin2X+2(sinX)^2]\/(1-tanX)=2[sinxcosx（cosx+sinx）]\/cosx-sinx=-28\/75 麻烦采纳，谢谢!
若cos(π\/4+x)=3\/5,17\/12π<x<7\/4π,求(sin2x+2(sinx)^2)\/(1-tanx)
又(17\/12)π<x<(7\/4)π ∴ cosx =（ - 根号2）\/10， sinx =（ - 7根号2）\/10 （4）同理可得：把（3）代入后解得结果与（4）相同 ∵（sin2x+2（sinx）^2）\/(1-tanx)=(2sinxcosx +2sin²x)\/[1-(sinx\/cosx)]把（4）代入得：原式= (7\/25 + 49\/25)\/(1-7...
cos(π\/4+x)=3\/5,17π\/12<x<7π\/4,求sin2x+2sin^2x\/1-tanx
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