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某参考资料中有这样一道习题 :已知 tan3 A、tan A是方程 x2 +6x +7=0的两根 ,求 2 sin2 A - cos4A - 12 cos2 A - cos4A - 1 的值 .解法 12 sin2 A - cos4A - 12 cos2 A - cos4A - 1 =- cos2 A - cos4Acos2 A - cos4A=- 2 cos Acos3 A2 sin Asin3 A =- 1tan A . tan3 A∵ ta
A reference contains the following exercise: It is known that tan3 A and tan A are the two equations x2 +6x +7=0, find the value of 2 sin2 A - cos4A - 12 cos2 A - cos4A - 1 . Solution 12 sin2 A - cos4A - 12 cos2 A - cos4A - 1 = - cos2 A - cos4Acos2 A - cos4A = - 2 cos Acos3 A2 sin Asin3 A = - 1tan A . tan3 A∵ ta