要完整既式
1)
tan@ + 1
sin@-sin2@ cos@(1+sin@)
2)
1 - 1 - cos2@
sin2@ tan2@
3)
Given that tan@ = 6 , find the value of 3cos@+4sin@
7 sin@ - 6cos@
F.3 (3條)急!!數 20點!!!
2010-09-20 6:14 am
回答 (2)
2010-09-20 8:45 pm
✔ 最佳答案
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參考: My Maths knowledge
2010-09-20 6:23 am
Q1 and Q2 are unreadable.............
Q3:
3cos@+4sin@/7sin@-6cos@
3+4(6)/7(6)-6
27/36
3/4
__________________
(3cos@+4sin@/cos@)/(7 sin@ - 6cos@ )/cos@)
Maybe I calculate wrong
Q3:
3cos@+4sin@/7sin@-6cos@
3+4(6)/7(6)-6
27/36
3/4
__________________
(3cos@+4sin@/cos@)/(7 sin@ - 6cos@ )/cos@)
Maybe I calculate wrong
收錄日期: 2021-04-13 17:32:26
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100919000051KK01978