高一数学题目有几道不清楚``帮下忙 (1) 已知tan^2α=tan^2β +1,求证cos(2α)+sin^2β =0(2) 已知x+(1/x)=2cos(π/24),求x^8+1/x^8(3) 求证cos^2α+cos^2(α-60度)+cos^2(α+60度)的值是与α无关的定值(4) 已知α,β为锐角.且3sin^2α+2sin^2β=1,3sin(2α)-2sin(2β)=0.求证:α+2β=π/2高手帮下忙``我还会追加的

2).x+(1/x)=2cos(π/24),
x^2+(1/x)^2=[x+(1/x)]^2-2=4cos^2(π/24)-2
=2cos(π/12)
x^4+1/x^4=[x^2+(1/x)^2]^2-2=4cos^2(π/12)-2
=2cos(π/6)
x^8+1/x^8=[x^4+1/x^4]^2-2=4cos^2(π/6)-2
= 2cos(π/3)=1.(x为复数)
3).cos^2α+cos^2(α-60度)+cos^2(α+60度)
=cos^2α+(cosαcos60+sinαsin60)^2+(cosαcos60-sinαsin60)^2
=cos^2α+2cos^2αcos^2(60)+2sin^2αsin^2(60)
=cos^2α+1/2cos^2α+3/2sin^2α=3/2(cos^2α+sin^2α)
=3/2,即得证
4)3sin^2α+2sin^2β=1,3sin^2α