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本文将首先利用余弦定理导出三个不等式。这些不等式用在求证某类(与三角形元素有关的)极值问题时是非常有趣的。若角A、B、C能组成一个三角形,且所对边为a、b、c,则有下列不等式: sin(A/2)≤(a/2(bc~(1/2)),sin(B/2)≤b/2(ca~(1/2)), sin(C/2)≤c/2(ab~(1/2))。证明:利用余弦定理和半角公式: ∵cosA=1-2sin~2(A/2)=(b~2+c~2-a~2)/2bc即 2sin~2(A/2)≤a~2/2bc (∵(b-c)~2≥0)于是,sin(A/2)≤a/2(bc~(1/2))(∵sin(A/2)>0, ∴取正号) 显然△ABC为等腰三角形时(b=c)取等号.同理有:sin(B/2)≤b/2(ac~(1/2)),sin(C/2)≤c/2(ab~(1/2)) 例1 任意三角形的三边长为a、b、c。求证:
This article will first use cosine theorem to derive three inequalities. These inequalities are very interesting when it comes to proving the extremum of a class (related to a triangle element). If the angles A, B, and C can form a triangle and the edges are a, b, and c, then there are the following inequalities: sin(A/2)≤(a/2(bc~(1/2)),sin (B/2)≤b/2(ca~(1/2)), sin(C/2)≤c/2(ab~(1/2)) Proof: Using cosine theorem and half-angle formula: ∵cosA =1-2sin~2(A/2)=(b~2+c~2-a~2)/2bc is 2sin~2(A/2)≤a~2/2bc (∵(bc)~2≥ 0) Then, sin(A/2)≤a/2(bc~(1/2))(∵sin(A/2)>0, get the positive sign) Obviously, △ABC is an isosceles triangle (b= c) Take the equal sign. The same holds for sin(B/2) ≤ b/2(ac~(1/2)) and sin(C/2)≤c/2(ab~(1/2)) 1 The three sides of an arbitrary triangle are a, b, and c.