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合分比定理(若a/b=c/d,则(a+b)/(a+b)=(c+d)/(c-d))在代数和几何方面的广泛应用,不少书刊中已作过阐述。但合分比定理在三角学中的应用,却谈得较少。其实,在证明三角恒等式或求值时,应用合分比定理常能简捷地得到答案。本文想通过以下几道例题进行说明。例1 tg~2α=1+2tg~2β,求证 cos~2β==1十cos2α。证先将已知条件变形为 (tg~2α)/1=(1+2tg~2β)/1,应用合分比定理得, (1-tg~2α)/(1+tg~2α)=(-2tg~2β)/(2(1+tg~2β)),而(1-tg~2α)/(1+tg~2α)=cos2α,(-2tg~2β)/2(1+tg~2β)=1/(1+tg~2β)-1=cos~2β-1,
Convergence ratio theorem (if a/b=c/d, (a+b)/(a+b)=(c+d)/(cd)) Widely used in algebra and geometry, many publications It has been elaborated. However, the application of the combination ratio theorem in trigonometry is less discussed. In fact, when we prove the trigonometric identities or the evaluations, the application of the fractional ratio theorem can often get the answer simply. This article wants to explain through the following examples. Example 1 tg~2α=1+2tg~2β, verification cos~2β==1 ten cos2α. The proof first transforms the known condition into (tg~2α)/1=(1+2tg~2β)/1, and applies the fractional ratio theorem. (1-tg~2α)/(1+tg~2α)=( -2tg~2β)/(2(1+tg~2β)), while (1-tg~2α)/(1+tg~2α)=cos2α, (-2tg~2β)/2(1+tg~2β )=1/(1+tg~2β)-1=cos~2β-1,