如图,在△ABC中,AC=2,BC=1,cosC=3/4. (1)求AB的值;(2)求sin(2A+C)的值.

问题描述:

如图,在△ABC中,AC=2,BC=1,cosC=

3
4


(1)求AB的值;
(2)求sin(2A+C)的值.

(1)由余弦定理,AB2=AC2+BC2-2AC•BC•cosC=4+1-2×2×1×

3
4
=2.
那么,AB=
2

(2) 由cosC=
3
4
,且0<C<π,
sinC=
1-cos2C
=
7
4
.由正弦定理,
AB
sinC
=
BC
sinA

解得sinA=
BCsinC
AB
=
14
8

所以,cosA=
5
2
8

由倍角公式sin2A=2sinA•cosA=
5
7
16

cos2A=1-2sin2A=
9
16

sin(2A+C)=sin2AcosC+cos2AsinC=
3
7
8