tanAtanB=1/根號(hào)3，求(2-cos2A)(2-cos2B)

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算出來了，占個(gè)位，要打字。解：根據(jù)萬能公式cos2α＝[1-(tgα平方)]/[1+(tgα平方) ] (2-cos2A)×(2-cos2B)＝{2－[1-(tgA平方)]/[1+(tgA平方) ]}×{2－[1-(tgB平方)]/[1+(tgB平方) ]}　考慮tgA符號(hào)的難寫，可以令tgA＝m,tgB＝n 另外，用^2表示平方上式就可以化為：＝{2-[(1-m^2)/(1+m^2)]}×{2－[(1-n^2)/(1+n^2)]}＝[(3m＾2+1)/(1+m^2)]×[(3n＾2+1)/(1+n^2)]＝(3m＾2+1)(3n＾2+1)/[(1+m^2)(1+n^2)]＝(9m^2×n^2+3m^2+3n^2+1)/(m^2×n^2+m^2+n^2+1)∵tanAtanB=1/根號(hào)3　∴mn＝1/根號(hào)3　∴(mn)平方＝1/3上式的(9m^2×n^2+3m^2+3n^2+1)/(m^2×n^2+m^2+n^2+1)＝(3+3m^2+3n^2+1)/(1/3+m^2+n^2+1)＝(4+3m^2+3n^2)/(4/3+m^2+n^2)＝(4+3m^2+3n^2)/[(1/3)*(4+3m^2+3n^2)]＝3∴(2-cos2A)×(2-cos2B)＝3。