高一二倍角三角函数1.求证：(sinx+cosx-1)(sinx-cosx+1)/sin2x=tanx/22.化简:（1+sin4x-cos4x)/(1+sin4x+cos4x) - (1+sin4x+cos4x)/(1+sin4x-cos4x)3.已知sin(π/4 -x)=5/13,0

高一二倍角三角函数1.求证：(sinx+cosx-1)(sinx-cosx+1)/sin2x=tanx/22.化简:（1+sin4x-cos4x)/(1+sin4x+cos4x) - (1+sin4x+cos4x)/(1+sin4x-cos4x)3.已知sin(π/4 -x)=5/13,0
高一二倍角三角函数
1.求证：(sinx+cosx-1)(sinx-cosx+1)/sin2x=tanx/2
2.化简:
（1+sin4x-cos4x)/(1+sin4x+cos4x) - (1+sin4x+cos4x)/(1+sin4x-cos4x)
3.已知sin(π/4 -x)=5/13,0

高一二倍角三角函数1.求证：(sinx+cosx-1)(sinx-cosx+1)/sin2x=tanx/22.化简:（1+sin4x-cos4x)/(1+sin4x+cos4x) - (1+sin4x+cos4x)/(1+sin4x-cos4x)3.已知sin(π/4 -x)=5/13,0
1.证明：(sinx+cosx-1)(sinx-cosx+1)/sin2x=tanx/2
(sinx+cosx-1)(sinx-cosx+1)/sin2x
=[sinx+(cosx-1)][sinx-(cosx-1)]/sin2x
=[(sinx)^2-(cosx-1)^2]/sin2x
=[(sinx)^2-(cosx)^2-1+2cosx]/sin2x
=[2cosx-2(cosx)^2]/sin2x
=2cosx(1-cosx)/2sinxcosx
=(1-cox)/sinx
=tanx/2 得证
2.化简:
（1+sin4x-cos4x)/(1+sin4x+cos4x) - (1+sin4x+cos4x)/(1+sin4x-cos4x) 通分
=[（1+sin4x-cos4x)^2+(1+sin4x+cos4x)^2]/[(1+sin4x+cos4x)*(1+sin4x-cos4x)]
=[1+(sin4x)^2+(cos4x)^2+2sin4x-2cos4x-2sin4xcos4x+1+(sin4x)^2+(cos4x)^2+2sin4x+2cos4x+2sin4xcos4x]/[(1+sin4x)^2-(cos4x)^2]
=[2+2(sin4x)^2+2(cos4x)^2+4sin4x]/[1+2sin4x+(sin4x)^2-(cos4x)^2]
=[2+2+4sin4x]/[2sin4x+2(sin4x)^2]
=4[1+sin4x]/[2sin4x(1+sin4x)]
=2/sin4x
3.已知sin(π/4 -x)=5/13,0

1.(sinx+cosx-1)(sinx-cosx+1)=sin2x-(cosx-1)^2
=sin^2x-cos^2x+2cosx-1
=2cosx(1-cosx)
(sinx+cosx-1)(sinx-cosx+1)/sin2x=2cosx(1-cosx)/(2sinxcosx)
=(1-cosx)/sinx=(2sin^2x/2)/2(sinx/2)(cosx...

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1.(sinx+cosx-1)(sinx-cosx+1)=sin2x-(cosx-1)^2
=sin^2x-cos^2x+2cosx-1
=2cosx(1-cosx)
(sinx+cosx-1)(sinx-cosx+1)/sin2x=2cosx(1-cosx)/(2sinxcosx)
=(1-cosx)/sinx=(2sin^2x/2)/2(sinx/2)(cosx/2)
=tanx/2
2.（1+sin4x-cos4x)/(1+sin4x+cos4x) =(2sin2xcos2x+2sin^22x)/(2sin2xcos2x+2cos^22x)
=tan2x
同理(1+sin4x+cos4x)/(1+sin4x-cos4x) =cot2x
原式=tan2x-cot2x=(sin^22x-cos^22x)/sin2xcos2x=-2cot4x

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(sinx+cosx-1)(sinx-cosx+1)/sin2x
= [sinx + (cosx-1)][sinx - (cosx-1)]/sin2x
= [(sinx)^2 - (cosx-1)^2]/sin2x
= [(sinx)^2 - (cosx)^2 + 2cosx - 1]/sin2x
= [(sinx)^2 - (cosx)^2 + 2cosx - (...

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(sinx+cosx-1)(sinx-cosx+1)/sin2x
= [sinx + (cosx-1)][sinx - (cosx-1)]/sin2x
= [(sinx)^2 - (cosx-1)^2]/sin2x
= [(sinx)^2 - (cosx)^2 + 2cosx - 1]/sin2x
= [(sinx)^2 - (cosx)^2 + 2cosx - (sinx)^2 - (cosx)^2]/sin2x
= [2cosx - 2(cosx)^2]/(2 sinx * cosx)
= (1 - cosx)/sinx
= [1 - 1 + 2(sinx/2)^2 ]/(2* sinx/2 * cosx/2)
= (sinx/2)/(cosx/2)
= tanx/2
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（1+sin4x-cos4x)/(1+sin4x+cos4x) - (1+sin4x+cos4x)/(1+sin4x-cos4x)
先对分母通分
= [(1+sin4x-cos4x)^2 - (1+sin4x+cos4x)^2]/[(1+sin4x+cos4x)(1+sin4x - cos4x)]
= (2+2sin4x)(-2cos4x)/[(1+sin4x+cos4x)(1+sin4x - cos4x)]
= -4(1+sin4x)*cos4x/[(1+sin4x)^2 -(cos4x)^2]
= -4(1+sin4x)*cos4x/[1 + 2sin4x + (sin4x)^2 - (cos4x)^2]
= -4(1+sin4x)*cos4x/[(sin4x)^2 + (cos4x)^2 + 2sin4x + (sin4x)^2 - (cos4x)^2]
= -4(1+sin4x)*cos4x/[2sin4x + 2(sin4x)^2]
= -2(1+sin4x)*cos4x/[sin4x(1+sin4x)]
= -2cos4x/sin4x
= -2 cot4x
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已知sin(π/4 -x)=5/13,0因为 (π/4 - x) + (π/4 + x) = π/2
所以
cos(π/4 + x) = sin(π/4 - x) = 5/13
cos2x = cos[(x+π/4) + (x-π/4)]
= cos(x+π/4)cos(x -π/4) - sin(x+π/4)sin(x-π/4)
其中 cos(x+π/4) 以及 sin(x-π/4) 已经已知 ,分别为 5/13 和 -5/13
下面求另外两项
利用 (sinA)^2 + (cosA)^2 = 1 以及 0可求出
cos(x-π/4) = 根号下{ 1 - [sin(x-π/4)]^2} = 12/13
sin(x+π/4) = 根号下{ 1 - [cos(x+π/4)]^2} = 12/13
因此
cos2x = cos(x+π/4)cos(x -π/4) - sin(x+π/4)sin(x-π/4)
= (5/13)*(12/13) - (12/13)*(-5/13)
= 120/169
原式
= cos2x/cos(π/4+x)
= (120/169)/(5/13)
= 24/13

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