在△ABC中，a,b,c分别是△ABC三个内角A,B,C的对边，且满足a+b=√3csinA+cc

Question

在△ABC中，a,b,c分别是△ABC三个内角A,B,C的对边，且满足a+b=√3csinA+ccosA①求角C的值②求cos^2A+cos^2B的取值范围 ，

Answer 1

①根据正弦定理，a/sinA=b/sinB=c/sinC
有(a+b)/c=(sinA+sinB)/sinC
根据题意，(sinA+sinB)/sinC=√3sinA+cosA
sinA+sin(A+C)=√3sinAsinC+cosAsinC

sinA+sinAcosC+cosAsinC=√3sinAsinC+cosAsinC
sinA(1+cosC)=√3sinAsinC
1+cosC=√3sinC
2sin(C-π/6)=1
sin(C-π/6)=1/2
C-π/6=π/6 （5π/6舍去）
C=π/3
②cos^2A+cos^2B
=(2cos^2A+2cos^2B)/2
=(1+cos2A+1+cos2B)/2
=1+(cos2A+cos2B)/2
根据和差化积，=1+cos(A+B)cos(A-B)
=1-cosCcos(A-B)
=1-1/2*cos(A-B)
因为A-B=π-C-2B=2π/3-2B，且0<B<2π/3
所以-2π/3<A-B<2π/3，-1/2<cos(A-B)<=1，-1/4<1/2*cos(A-B)<=1/2
1/2<=1-1/2*cos(A-B)<5/4
即1/2<=cos^2A+cos^2B<5/4