Question

【高中】【数学】求详细解答！谢谢！最好发图片来！！

Answer 1

(cosA)^2-(cosB)^2=√3sinAcosA-√3sinBcosB
1/2(1+cos2A)-1/2(1+cos2B)=√3/2sin2A-√3/2sin2B
cos2A-cos2B=√3(sin2A-sin2B) ,(等式两边和差化积得下)
-2sin(2A+2B)/2*sin(2A-2B)/2=2√3cos(2A+2B)/2*sin(2A-2B)/2
-sin(A+B)sin(A-B)=√3cos(A+B)sin(A-B)
sin(A-B)≠0
-sin(A+B)=√3cos(A+B)
-sinC=-√3cosC
tanC=√3
C=π/3
2) c=√3 ,C=π/3,B=C=2π/3-A
因2R=a/sinA=b/sinBc/sinC=√3/√3/2=2
a+b+c=√3+2RsinA+2RsinB
=√3+2R(sinA+sinB)
=√3+2R[sinA+sin(2π/3-A)]
=√3+2[sinA+√3/2cosA+1/2sinA]
=√3+2(3/2sinA+√3/2cosA)
=√3+2√3(√3/2sinA+1/2cosA)
=√3+2√3sin(A+π/6)
因A∈(0，2π/3)
A+π/6∈(2π/3,5π/6)
sin(A+π/6)在(2π/3,5π/6)上值域为:(1/2,√3/2)
a+b+c的值域为:(2√3,3+√3)
2√3 <a+b+c<3+√3

Answer 2

用三角关系式、余弦定理、基本不等式(见补充图片)

追答

Answer 3

Answer 4

图片太模糊