1)等比数列中a2=sina+cosa,
a3=1+sin2a=(sina)^2+(cosa)^2+2sinacosa=(sina+cosa)^2
所以，公比q=a3/a2=sina+cosa
由此an=(sina+cosa)^(n-1)
2sin2a-(1/2)cos4a+3/2
=2sin2a-(1/2)[1-2(sin2a)^2]+3/2
=(sin2a)^2+2sin2a+1
=(sin2a+1)^2
=(sina+cosa)^4
n-1=4--->n=5,所以，此项是数列的第5项
2)tan(pi-a)=4/3--->-tana=4/3--->tana=-4/3
(csa)^2=1/[1+(tana)^2]=1/[1+(-4/3)^2]=9/25
pi/2<a<pi--->cosa=-3/5,sina=4/5,sina+cosa=1/5.
--->an=(1/5)^(n-1),
Sn=[1-(1/5)^n]/(1-1/5)=(1/4)[5-(1/5)^(n-1)]
