在三角形ABC中,角A,B,C所对的边为a,b,c,已知sinC/2=根号10/4
问题描述：
在三角形ABC中,角A,B,C所对的边为a,b,c,已知sinC/2=根号10/4
若三角形ABC面积为3根号15/4且(sinA)^2+(sinB)^2=13/16(sinC)^2
（1）求a,b,c
(2)若a
答
a/sinA=b/sinB=c/sinC=k
(sinA)^2+(sinB)^2=(13/16)(sinC)^2
(ksinA)^2+(ksinB)^2=(13/16)(ksinC)^2
a^2+b^2=(13/16)c^2
2abcosC=a^2+b^2-c^2=(-3/16)c^2
abcosC=(-3/32)c^2
S=absinC/2=3√15/4,sin(C/2)=√10/4 cos(C/2)=√6/4,sinC=√15/4,cosC=2(cosC/2)^2-1=-1/4
ab=(3/2)*4=6
-6/4=(-3/32)c^2 c=4,
a^2+b^2=13
ab=6
a+b=5,a-b=1或b-a=1
a=3 b=2 或a=2 b=3
2
f(x)=根号bsinwx+(a-c)（cos wx/2）^2
=√3sinwx-2(coswx/2)^2=√3sinwx-coswx-1=2sin(wx-π/6)
sin(wx-π/6)=1/2 wx-π/6=π/6+2kπ或wx-π/6=5π/6+2kπ
w=2时,在[t,t+π)内有两个点sin(wx-π/6)=1/2
f(x)=2sin(2x-π/6)
-π/2+2kπ