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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
人气:253 ℃ 时间:2020-03-21 01:17:06
解答
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
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