Question

求答案 高一数学

Answer 1

1.B:原式=Cos（15+15）=√3/2
2.C:(cos2x)/[sin(x-π/4)]
=[(cosx)^2-(sinx)^2]/[((根号2)/2)sinx-((根号2)/2)cosx]
=(cosx-sinx)(cosx+sinx)/[-((根号2)/2)(cosx-sinx)]
=(cosx+sinx)/[-(根号2)/2]=-(根号2)/2,
所以 sinx+cosx= 1/2.
3.(SinQ+cosQ)^2=1+2SinQcosQ=1+根号(1-(Cos2Q)^2)=1/25==>Cos2Q=±0.28
又因为：Q∈【TT/2，3TT/4】故，2Q∈【TT，3TT/2】故Cos2Q<0==>Cos2Q=-0.28
4.由题意：a=b,C=5c=2a+c==>c=1/2a Cos顶角=(2a^2-1/4a^2 )/2a^2=7/8