设a∈R函数f(x)=cos x(asin x-cos x)+cos2(
+x)满足f(-
)=f(0).
(1)求f(x)的单调递减区间;
(2)设锐角△ABC的内角A、B、C所对的边分别为a、b、c,且
=
,求f(A)的取值范围.
设a∈R函数f(x)=cos x(asin x-cos x)+cos2(+x)满足f(-)=f(0).
(1)求f(x)的单调递减区间;
(2)设锐角△ABC的内角A、B、C所对的边分别为a、b、c,且=,求f(A)的取值范围.
解:(1)f(x)=cos x(asin x-cos x)+cos2(+x)
=
sin 2x-cos 2x,
由f(-)=f(0)得-
+
=-1,
∴a=2
,
∴f(x)=sin 2x-cos 2x=2sin(2x-
),
由2kπ+≤2x-≤2kπ+
π得kπ+≤x≤kπ+
π,k∈Z,
∴f(x)的单调递减区间为[kπ+,kπ+π].
(2)∵=,
由余弦定理得
=
=,
即2acos B-ccos B=bcos C,由正弦定理得
2sin Acos B-sin Ccos B=sin Bcos C,
2sin Acos B=sin(B+C)=sin A,cos B=,
∴B=,
∵△ABC为锐角三角形,
∴<A<,<2A-<
,
∴f(A)=2sin(2A-)的取值范围为(1,2].