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amaths

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http://i270.photobucket.com/albums/jj88/ucc041087/2.jpg help me to answer these Qs above .thz!~ 請寫清楚步驟，不要跳steps!!!!

最佳解答:

此文章來自奇摩知識+如有不便請留言告知

1)cos(3x/2) = cos(x/2) cos(3x/2)﹣cos(x/2) = 0 -2sin{[(3x/2)+(x/2)]/2} sin{[(3x/2)﹣(x/2)]/2} = 0 -2sin(2x/2) sin(x/2) = 0 -2sinx sin(x/2) = 0 sinx = 0 或 sin(x/2) = 0 sinx = 0°,180°,360° 或 sin(x/2) = 0°,180°,360° x = 0°,180°,360° 或 x = 0°,360°,720° x = 0°,180°,360°,720° 2)cosx + cos2x + cos3x = 0 cosx + cos3x + cos2x = 0 2cos[(x + 3x) / 2]cos[(x - 3x) / 2] + cos2x = 0 2cos2x cos(-x) + cos2x = 0 2cos2x cosx + cos2x = 0 cos2x(2cosx + 1) = 0 cos2x = 0 or cosx = -1 / 2 2x = 90°, 270°, 450°, 630° or x = 120°, 240° x = 45°, 135°, 225°, 315° or 120°, 240° 2008-02-20 21:57:24 補充： 根據公式：cosx﹣cosy = -2sin[(x+y)/2] sin[(x﹣y)/2]cosx+cosy = 2cos[(x+y)/2] cos[(x﹣y)/2]

其他解答:FAD2A23AB937987B
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