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在三角中,某些问题如我们能充分注意到它们的几何背景,并藉助于解析几何的有关知识,往往可以得到较为简洁的解法。本文列举数例,以资说明。例1 已知 cosa-cosβ=1/2,sina-sinβ=-1/3,求cos(a+β)。解:设x_1=cosa,y_1=sina;x_2=cosβ,y_2=sinβ。则可知点A(x_1,y_1),B(x_2,y_2)在单位圆x~2+y~2=1上。(图一) 又由(y_2-ly_1)/(x_2+x_1)=(sinβ-sina)/(cosβ-cosa)=(1/3)/(-1/2)=-2/3玄j 故直线AB的斜率为-2/3。设直线AB的方程为y=-2/3x+b,将此代入x~2+y~2=1并整理得13x~2-12bx+9(b~3-1)
In trigonometry, certain problems such as we can pay enough attention to their geometric background, and with the aid of the relevant knowledge of analytical geometry, we can often get a more concise solution. This article lists a few examples to illustrate. Example 1 It is known that cosa-cosβ=1/2, sina-sinβ=-1/3, find cos(a+β). Solution: Let x_1=cosa,y_1=sina;x_2=cosβ,y_2=sinβ. It can be seen that point A(x_1,y_1) and B(x_2,y_2) are on the unit circle x~2+y~2=1. (Fig. 1) Again from (y_2-ly_1)/(x_2+x_1)=(sinβ-sina)/(cosβ-cosa)=(1/3)/(-1/2)=-2/3 mysterious j The slope of line AB is -2/3. Let the equation of the line AB be y=-2/3x+b, and substitute it into x~2+y~2=1 and sort it to 13x~2-12bx+9(b~3-1).