Question

问几道高一数学题，求过程

1.求最简形式：sin2x/(1+cos2x）*cosx/(1+cosx)*sinx/(1-cosx)
2.化简：根号（2-sin平方2+cos4)
3.已知4sinx=1+cosx,x≠2kπ+π，k∈Z，则tanx=?
4.已知等腰三角形顶角A的正弦值为5/13，求底角B的余弦
5.已知cos(π+a)=4/5,cos(π/2-a)=3/5,那么点P（cos2a,sin2a)在第几象限？

Answer 1

1,sin2x/(1+cos2x）*cosx/(1+cosx)*sinx/(1-cosx)
=2sinx*cosx/[2(cosx)^2]*(cosx*sinx)/[(1+cosx)(1-cosx)]
=2sinx*cosx/[2(cosx)^2]*(cosx*sinx)/[1-(cosx)^2]
=2sinx*cosx/[2(cosx)^2]*(cosx*sinx)/[(sinx)^2]
=1

2,√[2-(sin2)^2+cos4]
=√[2-(sin2)^2+1-2(sin2)^2]
=√3[1-(sin2)^2]
=√3*|cos2|
=-√3*cos2. (cos2<0)

3,4sinx=1+cosx
8*sin(x/2)*cos(x/2)=2[cos(x/2)]^2
tan(x/2)=1/4 (x≠2kπ+π)
tanx=2tan(x/2)/{1+[tan(x/2)]^2}=8/17

4, A+2B=π
A=π-2B
sinA=5/13
sin(π-2B)=sin2B=5/13
2sinB*cosB=5/13
(sinB)^2+(cosB)^2=1
(sinB+cosB)^2=18/13
(sinB-cosB)^2=8/13
sinB+cosB=3√26/13 (B为等腰三角形内角，sinB>0,cosB>0)
sinB-cosB=2√26/13 ，或 sinB-cosB=-2√26/13
cosB=√26/26 ，或 cosB=5√26/26

5,cos(π+a)=4/5,cos(π/2-a)=3/5
cosa=-4/5, sina=3/5
cos2a=2(cosa)^2-1=7/25>0
sin2a=2sina*cosa=-24/25<0
点P(cos2a,sin2a)在第四象限