數學三角函數 求過程(急!!!

1.化簡(1+sinθ)/(1+cosθ)．(1+secθ)/(1+cscθ)得（Ａ）tanθ簡化計算令sx=sinθ,cx=cosθw1=[(1+sx)/(1+cx)]*[cx*sx(1+secx)/cx*sx(1+cscx)]=[(1+sx)/(1+cx)]*[(1+cx)sx/(1+sx)cx]=sx/cx=tanx
2.sinθ=(-2x+1)/x，x的範圍是？（Ａ）1/3<=x<=1-1<=sinθ<=1 => -1<=(-2x+1)/x<=1同時*x^2: -x^2<=(-2x+1)x<=x^2同時+x^2: 0<=-x^2+x<=2x^2左邊兩者: 0<=x(x-1) => 0<=x<=1右邊兩者: x(3x-1)>=0 => x<=0, x>=1/3兩者合併: 1/3<=x<=1.......ans



2014-03-21 07:31:33 補充：
3.tx=-3/4且cx<0，則５sx+4secx等於？（Ｃ）-2

tx=-3/4且cx<0 => 第2象限: sx=3/5, secx=-5/4

w3=5*3/5-4*5/4

=3-5

=-2

2014-03-21 07:39:43 補充：
4.ct(11π/4)+t(45π/4)+c(14π/4)+s(-29π/4)=?（Ａ）-1

=ct(2π+3π/4)+t(11π+π/4)+c(3π+π/2)-s(7π+π/4)

=ct(3π/4)-t(π/4)-c(π/2)+s(π/4)

=-1-1-0+1

=-1

2014-03-21 07:42:01 補充：
5.化簡c(180度+x)*c(180度-x)-s(180度+x)*s(180度-x)（Ｄ）1

=-c(x)*[-c(x)]-[-s(x)]*s(x)

=c^2(x)+s^2(x)

=1

2014-03-21 08:20:07 補充：
6.s^2(π/8)+s^3(π/8)+s^5(π/8)+s^7(π/8)=?（Ｂ）2

=0.211898

=\=2

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