设f(x)=6cos平方＋根号3sin2x,三角形abc中锐角a满足f(a)=3-2倍根号三,角b=pai/12,求（a/b+b/a)-c^2/ab值

题目

设f(x)=6cos平方＋根号3sin2x,三角形abc中锐角a满足f(a)=3-2倍根号三,角b=pai/12,求（a/b+b/a)-c^2/ab值

答案

f(x) = 6cos²x+√3 sin2x = 3 + 3cos2x + √3 sin2x而f(A) = 3 - 2√3化简得√3/2 cos2A + 1/2 sin2A = -1 ---> sin(2A + 60°) = -1解得A = 60°而B = 15°则C = 105° = 45° + 60°a/b + b/a -c²/ab = ...