|
ÊÁѤÃÊÁÒªÔ¡ | ¤ÙèÁ×Í¡ÒÃãªé | ÃÒª×èÍÊÁÒªÔ¡ | »¯Ô·Ô¹ | ¢éͤÇÒÁÇѹ¹Õé | ¤é¹ËÒ |
|
à¤Ã×èͧÁ×ͧ͢ËÑÇ¢éÍ | ¤é¹ËÒã¹ËÑÇ¢é͹Õé |
#1
|
||||
|
||||
µÃÕ⡳«Ñ¡¹Ô´¤ÃѺ
1)¨§·Ó $\cos{7A}\bullet \cos{3A}+\sin{7A}\bullet \sin{3A}$ ãËéàËÅ×;¨¹ìà´ÕÂÇ
2)¶éÒ $\cos({A+B})=\frac{4}{5}$ áÅÐ $\sin{A}=\frac{5}{13}$ àÁ×èÍÁØÁ $A+B$à»ç¹ÁØÁáËÅÁ áÅéǨ§ËÒ¤èÒ $\tan{B}$
__________________
I think you're better than you think you are. |
#2
|
||||
|
||||
µÍº¢éÍ 1 ¡è͹¹Ð¤ÃѺ ¨Ñ´ãËéÍÂÙèã¹ÃÙ» cos(a-b) ¨Ðä´éÇèÒ
cos7A•cos3A+sin7A•sin3A = cos(7a-3a)=cos4a¤ÃѺ (äÁèá¹èã¨ÇèÒãËé·ÓẺ¹ÕéËÃ×Í»ÅèÒÇáµèÁѹ¡ç¾¨¹ìà´ÕÂÇÍèÐ) ¢éÍ 2 ÍÂÒ¡ãËéÁͧÇèÒ sin(A)=sin((A+B)-A) cos(A+B)=4/5 cosAcosB-sinAsinB=4/5 (á·¹¤èÒ) 12cosB-5sinB=52/5 ÊÁ¡Òà 1 sin((A+B)-A)=sin(A+B)cosA-cos(A+B)sinA (á·¹¤èÒ) 12cosB-16sinB=100/3 ÊÁ¡Òà 2 á¡éÍÍ¡ÁÒ¨Ðä´é tanB= 16/63 Åͧ·Ó´Ù¹Ð¤ÃѺ 28 ÊÔ§ËÒ¤Á 2008 22:46 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ nongtum à˵ؼÅ: double post |
#3
|
||||
|
||||
¤Ø³ [SIL] à¡è§ÁÒ¡àÅ ¼Á¢Íá¶ÁÇÔ¸Õá¡é¢éÍÊͧẺãªéÃÙ»ªèÇÂà¾ÔèÁàµÔÁ¤ÃѺ (µÍ¹¹ÕéªÍºÇÒ´ÃÙ»)
|
#4
|
||||
|
||||
¨êÒ¡ææ ¾Õè Purriwatt ªÁ ¤Ó¶ÒÁ¢éÍáá·Õè¼ÁµÍºã¹ M.C àŹФÃѺà¹ÕèÂ
|
|
|