SUPER EASY !!EASY MATHS (好趕ga)
2010-11-09 1:34 am
圖片參考：http://imgcld.yimg.com/8/n/HA06682457/o/701011080092413873409910.jpg
QUICK!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
SHow all workings, even the simple one!
anyone who helped me this time, i will help u in the 1st priority in Econ!!!
回答 (3)
2010-11-09 1:45 am
✔ 最佳答案
20(cos334 + isin334)/15(cos199 + isin199)= 4(cos334 + isin334)[cos(-199) + isin(-199)]/3
= (4/3)[cos(334 - 199) + isin(334 - 199)]
= (4/3)(cos135 + isin135)
= (4/3)(-1/root2 + i/root2)
= -2root2/3 + 2root2i/3
8(cospi/6 + isinpi/6)/2(cospi/6 - isinpi/6)
= 4(cospi/6 + isinpi/6)/[cos(-pi/6) + isin(-pi/6)]
= 4(cospi/6 + isinpi/6)(cospi/6 + isinpi/6)
= 4[cos(pi/6 + pi/6) + isin(pi/6 + pi/6)]
= 4(cospi/3 + isinpi/3)
= 2 + 2root3i
2010-11-08 17:54:59 補充：
第2條既答案有問題,似乎唔係-1/3i
參考: me
2010-11-09 2:22 am
This is the way i understand:
=4(cospi/6+pi/6)+isin(pi/6+pi/6)
=4(cospi/3+isinpi/3)
=4(1/2+root3/2i)
=2+2root3i.
Your ans is correct!
=4(cospi/6+pi/6)+isin(pi/6+pi/6)
=4(cospi/3+isinpi/3)
=4(1/2+root3/2i)
=2+2root3i.
Your ans is correct!
收錄日期: 2021-04-23 20:48:53
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20101108000051KK00924