Question

在三角形ABC中，角A,B,C所对的边分别为a,b,c,已知a=1,b=2,cosc=1/4.求cos(A-C).已知c=2

Answer 1

由余弦定理
c²=a²+b²-2abcosC
=1+4-2×1×2×1/4
=5-1
=4
c=2
cosA=(b²+c²-a²)/2bc
=(4+4-1)/2×2×2
=7/8
sinA=√15/8
cosC=1/4
sinC=√15/4
cos(A-C).
=cosAcosC+sinAsinC
=(7/8)×(1/4)+(√15/8×（√15/4）
=7/32+15/32
=22/32
=11/16

Answer 2

cosA=（b²+c²-a²）/（2bc)=3/8
sinA=根号55/8
cosc=1/4,则sinc=根号15/4
cos(A-C)=cosAcosC+sinAsinC=3/8*1/4-根号55/8*根号15/4=(3-5根号33）/32