Question

2(sin²A+sin²B)＝sin²C，tanC＝2，求＜A

Answer 1

sinB+sinA=3(sin2A-sin2B)/2（sinB-sinA）
2(sinB+sinA)（sinB-sinA）=3(sin2A-sin2B)
2(sinB)^2-2(sinA)^2=3(sin2A-sin2B)
(1-cos2B)-(1-cos2A)=3(sin2A-sin2B)
cos2A-cos2B=3(sin2A-sin2B) [ 等式两边和差化积得]
-2sin(2A+2B)/2*sin(2A-2B)/2=3*2cos(2A+2B)/2*sin(2A-2B)/2
-sin(A+B)*sin(A-B)=3*cos(A+B)*sin(A-B)
sinC*sin(A-B)=3*cosC*sin(A-B)
sin(A-B)(sinC-3cosC)=0
sin(A-B)=0 或 sinC-3cosC=0
A=B 或 tanC=3
C=arc tan3