设△ABC的内角A,B,C的对边长分别为a,b,c,已知a,b,c成等比数列1)求证a^2+b^2+c^2>(a-b+c)^2,(2)若cos(A-C)+cosB=3/2,求B
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![设△ABC的内角A,B,C的对边长分别为a,b,c,已知a,b,c成等比数列1)求证a^2+b^2+c^2>(a-b+c)^2,(2)若cos(A-C)+cosB=3/2,求B](/uploads/image/z/5189043-3-3.jpg?t=%E8%AE%BE%E2%96%B3ABC%E7%9A%84%E5%86%85%E8%A7%92A%2CB%2CC%E7%9A%84%E5%AF%B9%E8%BE%B9%E9%95%BF%E5%88%86%E5%88%AB%E4%B8%BAa%2Cb%2Cc%2C%E5%B7%B2%E7%9F%A5a%2Cb%2Cc%E6%88%90%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%971%29%E6%B1%82%E8%AF%81a%5E2%2Bb%5E2%2Bc%5E2%3E%28a-b%2Bc%29%5E2%2C%282%29%E8%8B%A5cos%28A-C%29%2BcosB%3D3%2F2%2C%E6%B1%82B)
设△ABC的内角A,B,C的对边长分别为a,b,c,已知a,b,c成等比数列1)求证a^2+b^2+c^2>(a-b+c)^2,(2)若cos(A-C)+cosB=3/2,求B
设△ABC的内角A,B,C的对边长分别为a,b,c,已知a,b,c成等比数列
1)求证a^2+b^2+c^2>(a-b+c)^2,(2)若cos(A-C)+cosB=3/2,求B
设△ABC的内角A,B,C的对边长分别为a,b,c,已知a,b,c成等比数列1)求证a^2+b^2+c^2>(a-b+c)^2,(2)若cos(A-C)+cosB=3/2,求B
1)要证a^2+b^2+c^2>(a-b+c)^2
只需证a^2+b^2+c^2>a^2+b^2+c^2-(2ab+2bc-2ac)
只需证2ab+2bc-2ac>0,ab+bc-ac>0
因为b^2=ac 所以只需证b(a+c)-b^2>0
只需证b(a+c-b)>0
只需证a+c-b>0 (三角形两边之和大于第三边)显然成立
所以a^2+b^2+c^2>(a-b+c)^2成立
2) ccos (A-C) + cos B = 3/2
所以 cos(A-C) + cos(π - A - C) = 3/2
cos(A-C) + cos(π - A - C) = cos(A - C ) - cos(A + C)
= cosAcosC + sinAsinC - cosAcosC + sinAsinC = 2sinAsinC = 3/2
所以 sinAsinC = 3/4
正弦定理
a/sinA = b/sinB =c/sinC,而b^2 = ac,即 b/a = c/b
所以 sinB/sinA = sinC/sinB,所以 sinB ^2 = sinAsinc
所以sinB^2 = 3/4,sinB = 根号3/2,B=60°或120°
但是,如果 B=60°,则cosB = -1/2
带入 cos(A-C)+cosB=3/2,cos(A-C) = 2,矛盾,所以舍去这个值.