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Old 02 µØÅÒ¤Á 2006, 13:36
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Thumbs up ͹ءÃÁ ¢Í§ sin áÅÐ cos

ËÇÑ´´Õ¤Ñº ¼ÙéÃÙé·Ø¡·èÒ¹ ËÒÂ仹ҹàÅ ªèǧ¹Õé§Ò¹àÂÍШѧ äÁè¤èÍÂÁÕàÇÅÒ¤Ô´àÅ¢àÅÂ

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áÎÐ áÅÐÂѧ¤Ô´ÇÔ¸Õ¡ÒÃÍÔ¹·Ôà¡Ã·áººãËÁè ¡ç¤Ô´ÁÒáÅéÇ áµèÁÒÃÙé·ÕËÅѧÇèÒ¼Ô´

àÅÂàÅÔ¡ÅéÁ 令ԴÍÂèÒ§Í×è¹á·¹¤ÃѺ ¨¹µÍ¹¹Õé¼ÁÊÒÁÒö ËÒ͹ءÃÁ ¢Í§ sin áÅÐ cos ä´éáÅéÇ

¤ÃÒǹÕé¶Ù¡ªÑÇÃìæ¤ÃѺ à¾ÃÒÐÅͧ¡´à¤Ã×èͧ¤Ô´àÅ¢µÃǨÊͺ´ÙáÅéÇ áµè¼Á¡ÅÑÇÇèÒÍÒ¨¨ÐÁÕ¤¹¤Ô´ÁÒä´é¡è͹

áÅéǹèФÃѺ ÍÂèÒ§àªè¹ ¨§ËÒ¤èҢͧ

cos 1 + cos 2 + cos 3 + ....... + cos 90 = ?

sin 1 + sin 2 + sin 3 + ....... + sin 90 = ?

àÅÂÍÂÒ¡¶ÒÁ·èÒ¹¼ÙéÃÙéã¹·Õè¹ÕéÇèÒà¤ÂÁÕã¤Ã¤Ô´ËÒ͹ءÃÁ sin áÅÐ cos ä´éÁÒ¡è͹¼ÁÃÖÂѧ¤ÃѺ ¶éÒÁÕªèÇÂ

ÁҺ͡˹èÍÂ

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µÍº¾ÃéÍÁÍéÒ§ÍÔ§¢éͤÇÒÁ¹Õé
  #2  
Old 02 µØÅÒ¤Á 2006, 14:33
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͹ءÃÁ·Õè¤Ø³¡Ó˹´ÁÒ ÁÕº·¤ÇÒÁÍÂÙèáÅéÇã¹àÇ纹Õé

áµè ¶éҤسªèÇ·Ӣé͹Õé¨Ð¢Íº¤Ø³ÁÒ¡¤ÃѺ

Evaluate
$$ \cos1+\frac{\cos2}{2}+\frac{\cos3}{3}+... $$
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µÍº¾ÃéÍÁÍéÒ§ÍÔ§¢éͤÇÒÁ¹Õé
  #3  
Old 02 µØÅÒ¤Á 2006, 15:45
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ËÒ!!! ÁÕ¤¹¤Ô´ä´é¡è͹«ÐáÃéÇàËÃͤѺ

¶éÒ§Ñ鹢ͼÁ´Ùº·¤ÇÒÁ¹Ñé¹ä´éÁÑê¤ÃѺ ¨Ð´ÙÇèÒ¤Ô´àËÁ×͹¡Ñ¹ÃÖ»èÒÇ Êèǹ⨷Âì¢Í§¤Ø³ Mastermander

µÍ¹àÅÔ¡§Ò¹¼Á¨Ðà¡çºä»¤Ô´¤ÃѺ ( ⨷Âì ¤Ø³ Mastermander ÂѧÂÒ¡àËÁ×͹à´ÔÁ¹Ð

¤Ñº )
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¤ÇÒÁ¾ÂÒÂÒÁáÅÐÍ´·¹ à»ç¹¤Ø³ÊÁºÑµÔ˹Ö觢ͧ¹Ñ¡¤³ÔµÈÒʵÃì......

´Ù¼Á¨Ô ¼Á¾ÂÒÂÒÁ ËÒÊٵ÷ÑèÇä»ã¹¡ÒÃËÒÃÒ¡¢Í§ÊÁ¡ÒáÓÅѧ 5 ÍÂÙè¤ÃÖè§»Õ ÂѧäÁèä´éàŤѺ Î×Íææææ
µÍº¾ÃéÍÁÍéÒ§ÍÔ§¢éͤÇÒÁ¹Õé
  #4  
Old 02 µØÅÒ¤Á 2006, 16:04
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º·¤ÇÒÁâ´Â : ¹Ò ¾§Èì·Í§ á«èàÎé§


á¶Á⨷ÂìÍÕ¡ 2 ¢éͤÃѺ

1. ¨§ËÒ¤èÒ x ¨Ò¡ÊÁ¡ÒÃ
$$ \sum_{\theta=1}^{999}\sin(x\theta)=\sum_{\theta=1}^{999}\cos(x\theta) $$

2.
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02 µØÅÒ¤Á 2006 16:11 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ Mastermander
µÍº¾ÃéÍÁÍéÒ§ÍÔ§¢éͤÇÒÁ¹Õé
  #5  
Old 03 µØÅÒ¤Á 2006, 21:25
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âÅ¡¹ÕéÁÕ¤¹ÍÂÙè 10 »ÃÐàÀ· ¤×Í ¤¹·Õèà¢éÒã¨àÅ¢°Ò¹Êͧ áÅФ¹·ÕèäÁèà¢éÒã¨
µÍº¾ÃéÍÁÍéÒ§ÍÔ§¢éͤÇÒÁ¹Õé
  #6  
Old 04 µØÅÒ¤Á 2006, 10:58
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ÍéÒÇ ¾Õè¡Ã¤Ô´ä´é¡è͹áÅéÇàËÃͤѺ ÇÔ¸Õ¤Ô´µèÒ§¡Ñ¹àÅ硹éÍÂáµè¤ÓµÍºàËÁ×͹¡Ñ¹àŤѺ

µÍ¹¹Õé¼Á¾ÂÒÂÒÁËÒ͹ءÃÁ¢Í§ tan ÍÂÙè ÂѧËÒäÁèä´éàŤѺ àÎéÍ


¢Íº¤Ø³ ¤Ø³ Mastermander ·ÕèãËé¢éÍÁÙŹФѺ ⨷Âì 2 ¢éÍËÅѧ ¾ÍÃÙéá¹Ç㹡Ò÷ÓáÅéǤѺ ¶éÒÃÙé

¡ÒÃËÒ͹ءÃÁ sin áÅÐ cos ¡ç¤§·Óä´é áµè¢éÍáá·ÕèãËéÁÒ cos1 + (cos2)/2 + (cos3)/3 + .... ¹Õè·ÓÂѧä§

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¤Ø³ Mastermander ¹Õè¢éÍÁÙŤÇÒÁÃÙéá¹è¹»Öé¡áºº¹Õé ÁÕ˹ѧÊ×Í´Õæá¹Ð¹ÓºéÒ§ÁÑê¤Ѻ µÍ¹¹ÕéÍÂÒ¡

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Old 04 µØÅÒ¤Á 2006, 17:37
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I know only the answer is $-\frac{\ln(2-2\cos1)}{2}=0.0420195$...(by program)
Solution in next reply
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08 µØÅÒ¤Á 2006 22:17 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 2 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ Mastermander
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  #8  
Old 04 µØÅÒ¤Á 2006, 22:39
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·Õè¨ÃÔ§ÁÕáËÅè§ÃÇÁ E-book Ẻ˹ѧÊ×Íà»ç¹àÅèÁ æ ÁËÒÈÒÅÍÂÙè·Õè 2 ·Õè¹Ð¤ÃѺ à»ç¹áºº·ÕèÁѹËÁ´ÍÒÂØ·Ò§ÅÔ¢ÊÔ·¸Ôì»ÃÐÁÒ³¹Ñé¹ à¤ÂµÍºäÇéã¹ÊÑ¡¡ÃзÙé˹Öè§
µÍº¾ÃéÍÁÍéÒ§ÍÔ§¢éͤÇÒÁ¹Õé
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Old 08 µØÅÒ¤Á 2006, 22:14
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Short solution
(Editied)
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09 µØÅÒ¤Á 2006 13:44 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ Mastermander
µÍº¾ÃéÍÁÍéÒ§ÍÔ§¢éͤÇÒÁ¹Õé
  #10  
Old 09 µØÅÒ¤Á 2006, 10:12
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à»ç¹¡Òà solve ·ÕèàÂÕèÂÁ¤Ñº ÊÑé¹æ ä´é㨤ÇÒÁ
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µÍº¾ÃéÍÁÍéÒ§ÍÔ§¢éͤÇÒÁ¹Õé
  #11  
Old 09 µØÅÒ¤Á 2006, 19:55
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ÊÃØ»ä´éÇèÒ

$$\sin z+\frac{\sin 2z}{2}+\frac{\sin 3z}{3}+...=\arctan(\frac{\sin z}{1-\cos z})$$
$$\cos z+\frac{\cos 2z}{2}+\frac{\cos 3z}{3}+...=-\frac{\ln(2-2\cos z)}{2}$$


áµè $\frac{\sin z}{1-\cos z} = \cot\frac{z}{2}=\tan(\frac{\pi}{2}-\frac{z}{2})$

´Ñ§¹Ñ鹨֧ä´éÊÙµÃÊÇÂæ§ÒÁ·ÕèàË繴ѧ¹Õé

$$ \sum_{n=1}^{\infty} \frac{\sin nz}{n}=\arctan\big(\cot\frac{z}{2}\big)=\arctan\big(\tan(\frac{\pi}{2}-\frac{z}{2})\big)$$
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09 µØÅÒ¤Á 2006 20:20 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 2 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ Mastermander
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