cos(x-Ï/4)=cos[(x+Ï/4)-Ï/2]=sin(x+Ï/4)=-â[1-sin�0�5(x+Ï/4)]=-â[1-(3/5)�0�5]=-4/5
sin(2x)=-cos(2x+Ï/2)=-cos[2(x+Ï/4)]=1-2cos�0�5(x+Ï/4)=1-2�6�1(3/5)�0�5=7/25
[sin(2x)+2sin�0�5x]/(1-tanx)
=2(sinxcosx+sin�0�5x)/(1-sinx/cosx)
=2(cosx+sinx)/(1/sinx-1/cosx)
=2(cosx+sinx)sinxcosx/(cosx-sinx)
=cos(x-Ï/4)sin(2x)/cos(x+Ï/4)
=-4/5�6�17/25/(3/5)
=-28/75
已知cos(π\/4+x)=3\/5,17π\/12<x小于7π\/4,求sin2x+sin^2x\/1-tanx
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-sin�0�5(x+π\/4)]=-√[1-(3\/5)�0�5]=-4\/5 sin(2x)=-cos(2x+π\/2)=-cos[2(x+π\/4)]=1-2cos�0�...
cos(π\/4+x)=3\/5,17π\/12<x<7π\/4,求sin2x+2sin^2x\/1-tanx
因17π\/12<x<7π\/4，则17π\/6<2x<7π\/2，2x在二、三象限内，则cos2x=－24\/25 则：原式=[7\/25][1＋7\/25]\/[－24\/25]=－28\/3
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...
已知cos(π\/4+x)=3\/5,且7π\/12<x<7π\/4,求(2sin^x+sin2x)\/1-tanx.
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...
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 ...
若cos(π\/4+x)=3\/5,17\/12π<x<7\/4π,求(sin2x+2(sinx)^2)\/(1-tanx)
解得：cosx=（7*根号2）\/10 或 cosx =（ - 根号2）\/10 代入（1）得：sinx=（根号2）\/10 或 sinx =（ - 7根号2）\/10 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...
若cos(π\/4+x)=3\/5,17π\/12<x<7π\/4,求sin2x+2cos²x\/1-tanx
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 相应地代入式子即可得到结果 ...
cos(π\/4+x)=3\/5,17π\/12<x<7π\/4,求sin2x+2sin^2x\/1-tan^2x
sin 2x= 7\/25 ; cos[π\/4+(π\/4+x)] = cosπ\/4* cos(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, ...
若cos(π\/4+x)=3\/5 17π\/12<x<7π\/4求(sin2x+2cos平方x)\/(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^...
