求采纳
若cos(π\/4+x)=3\/5,17π\/12<x <7π\/4,求(sin2x+2cosx)\/1-tanx的值
sinx+cosx=√2sin(x+π\/4) 5π\/4<x+π\/4<2π 第四象限 所以sinx+cosx<0 (sinx+cosx)=sinx+cosx+2sinxcosx=1+7\/25=32\/25 sinx+cosx=-4√2\/5 原式=(2sinxcosx+2sinx)\/(1-sinx\/cosx) =2sinx(cosx+sinx)\/[(cosx-sinx)\/cosx] =2sinxcosx(cosx+sinx)\/(cosx-sinx) =-28...
高中数学,已知,cos(π\/4+x)=3\/5,17π\/12<x<7π\/4,求sin2x+2sin∧
17π\/12＜x＜7π\/4，得5π\/3＜x+π\/4＜2π 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\/2...
...+x)=3\/5,17π\/12<x<7π\/4,求(sin2x+2sin�0�5x)\/1-tanx的值...
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si...
若cos(π\/4+x)=3\/5 17π\/12<x<7π\/4求(sin2x+2cos平方x)\/(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
cos(π\/4+x)=3\/5,17π\/12<x<7π\/4,求sin2x+2sin^2x\/1-tanx
原式=[2sinxcosx＋2sin²x]\/[(cosx－sinx)\/cosx]=[2sinxcosx(sinx＋cosx)]\/[cosx－sinx]=[sin2x][(sinx＋cosx)²]\/[cos²x－sin²x]=[sin2x][1＋sin2x]\/[cos2x]sin2x=－cos[π\/2＋2x]=－cos[2(x＋π\/4)]=－[2cos²(π\/4＋x)－1]=7\/25 ...
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+2sinx)\/(1-
(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 ...
已知cos(π\/4+x)=3\/5,17π\/12<x<7π\/4,求sin2x+2sin^2x\/1-tanx
(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 ...
已知cos(π\/4+x)=3\/5,17π\/12<x<7π\/4,求(sin2x+2sin²x)\/1-
解 ∴[sin(2x)+2(sinx)^2]\/(1-tanx)=[2sinxcosx+2(sinx)^2]\/(1-sinx\/cosx)=[2sinx(cosx)^2+2(sinx)^2(cosx)]\/(cosx-sinx)=2sinxcosx(sinx+cosx)\/(cosx-sinx)=sin2x(sin(x+π\/4))\/(cos（x+π\/4）=-cos(π\/2+2x)(sin(x+π\/4))\/(cos（x+π\/4）=-(2cos^2...
cos(π\/4+x)=3\/5,17π\/12<x<7π\/4,求sin2x+2sin^2x\/1-tan^2x
π\/4+x)- sin π\/4*sin[(π\/4+x)]= 1\/2 (2^(1\/2)* (3\/5) - 1\/2 (2^(1\/2)* (-4\/5) = 7* 2^(1\/2)\/10 = cos (π\/2+x) = - sin x, sin x= -7* 2^(1\/2)\/10, tan x = 1\/7, finally, answer is :(7\/25) + 2* (49\/625)\/ [1- (1\/49)]
