Question

已知函数f（x）=2cosxsin（x+π/3）-√3/2 1.求函数f（x）的最小正周期T 2.若△ABC的三边a，b，c

Answer 1

1.
f(x)=2cosxsin（x+π/3）-√3/2

=
2cosx(sinx
cosπ/3
+
cosx
sinπ/3
)
-√3/2

=
cosx
sinx
+
√3cos²x
-√3/2

=
1/2
sin2x
+
√3/2
cos2x

=
sin(2x+π/3)
所以
T
=
2π/2
=
π
2.
cosB
=
(a²+
c²
-
b²)/(2ac)
(注意到b²=ac)

=
(a²
+
c²)/(2ac)
-
1/2

<=
1
-
1/2
=
1/2
所以，B>=
π/3
f(x)
=
sin(2x+π/3),
因为π
>
B>=π/3，所以
7π/3
>
2B+π/3
>=
π
所以f(B)的最大值为
√3/2。