已知4sin^2 x – 6sinx – cos^2 x + 3cosx = 0 ,求 cos(2x +π/4) / [(1-co_数学解答

已知4sin^2 x – 6sinx – cos^2 x + 3cosx = 0 ,求 cos(2x +π/4) / [(1-cos2x)(
已知4sin^2 x – 6sinx =cos^2 x -3cosx ,求 cos(2x +π/4) / (1-cos2x)(1-tan2x)的值

人气：348 ℃　时间：2019-09-03 10:34:36

解答

四分之三根号二,待求的式子展开化简,等式配成完全平方就可以.可能会算错,但思路不会错�Ҿ��ǻ��2sinx-3/2)^2-(cosx-3/2)^2=0Ҫ��ʽ�Ӱѷ��Ӻͷ�ĸ��ù�ʽչ��1-cos2x=2sin^2x,1-tan2x=1-sin2x/cos2x=(cos2x-sin2x)/cos2x,��ֱ��ù�ʽչ��Ȼ��ͻᷢ��cos2x-sin2x��ǹ�ʽӦ��ѧ��˰�><