设f(x)=6cos2x-3sin2x.(Ⅰ)求f(x)的最大值及最小正周期;(Ⅱ)△ABC中锐角A满足f(A)=3-23,B=π12,角A、B、C的对边分别为a,b,c,求(ab+ba)-c2ab的

题目简介

设f(x)=6cos2x-3sin2x.(Ⅰ)求f(x)的最大值及最小正周期;(Ⅱ)△ABC中锐角A满足f(A)=3-23,B=π12,角A、B、C的对边分别为a,b,c,求(ab+ba)-c2ab的

题目详情

设f(x)=6cos2x-
3
sin2x.
(Ⅰ)求f(x)的最大值及最小正周期;
(Ⅱ)△ABC中锐角A满足f(A)=3-2
3
B=
π
12
,角A、B、C的对边分别为a,b,c,求(
a
b
+
b
a
)-
c2
ab
的值.
题型:解答题难度:中档来源:不详

答案

(Ⅰ)f(x)=6cos2x-
3
sin2x
=6×class="stub"1+cos2x
2
-
3
sin2x
=3cos2x-
3
sin2x+3
=2
3
3
2
cos2x-class="stub"1
2
sin2x)+3
=2
3
cos(2x+class="stub"π
6
)+3,
∵-1≤cos(2x+class="stub"π
6
)≤1,
∴f(x)的最大值为2
3
+3;
又ω=2,∴最小正周期T=class="stub"2π
2
=π;
(Ⅱ)由f(A)=3-2
3
得:2
3
cos(2A+class="stub"π
6
)+3=3-2
3

∴cos(2A+class="stub"π
6
)=-1,
又0<A<class="stub"π
2
,∴class="stub"π
6
<2A+class="stub"π
6
class="stub"7π
6

∴2A+class="stub"π
6
=π,即A=class="stub"5
12

又B=class="stub"π
12
,∴C=class="stub"π
2

∴cosC=
a2+b2-c2
2ab
=0,
则(class="stub"a
b
+class="stub"b
a
)-
c2
ab
=
a2+b2-c2
ab
=2×
a2+b2-c2
2ab
=2cosC=0.

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