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  #1  
Old 26 µØÅÒ¤Á 2014, 19:52
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¨§ËÒ¤èҢͧ
$$(\frac{1}{sin(20)})^2+(\frac{1}{sin(40)})^2+(\frac{1}{sin(80) })^2 $$

»Å.ÁØÁà»ç¹Í§ÈÒ
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  #2  
Old 26 µØÅÒ¤Á 2014, 19:53
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Íѹ¹ÕéÇÔ¸Õ¼Á¹Ð¤ÃѺ ÁÕÇÔ¸Õä˹ÊÑ鹡ÇèÒ¹Õéáºè§»Ñ¹´éǤÃѺ
¨Ò¡ $sin(x)^2 = \frac{1-cos(2x)}{2}$

$$(\frac{1}{sin(20)})^2+(\frac{1}{sin(40})^2+(\frac{1}{sin(80) })^2 $$

$$ = \frac{2}{1-cos(40)}+\frac{2}{1-cos(80)}+\frac{2}{1-cos(160)}$$

$$ = 2*\frac{(1-cos(40))(1-cos(80))+(1-cos(40))(1-cos(160))+(1-cos(160))(1-cos(80))}{(1-cos(40))(1-cos(80)(1-cos(160)}$$

$$ =2*\frac{3-2(cos(40)+cos(80)+(cos(160))+cos(40)cos(80)+cos(40)cos(160)+cos(80)cos(160)}{1-(cos(40)+cos(80)+cos(160))+cos(40)cos(80)+cos(40)cos(160)+cos(80)cos(160)-cos(40)cos(80)cos(160)} $$

$cos(40)+cos(80)+cos(160)=0$

$$ =2*\frac{3+cos(40)cos(80)+cos(40)cos(160)+cos(80)cos(160)}{1+cos(40)cos(80)+cos(40)cos(160)+cos(80)cos(160)-cos(40)cos(80)cos(160)} $$

$cos(40)cos(80)+cos(40)cos(160)+cos(80)cos(160)=\frac{-3}{4}$
$cos(40)cos(80)cos(160)=\frac{-1}{8}$

¨Ðä´éÇèÒ

$$ =2*\frac{3+cos(40)cos(80)+cos(40)cos(160)+cos(80)cos(160)}{1+cos(40)cos(80)+cos(40)cos(160)+cos(80)cos(160)-cos(40)cos(80)cos(160)}$$

$$= 2*\frac{3-\frac{3}{4}}{1-\frac{3}{4}+\frac{1}{8}} $$

$$=12$$
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26 µØÅÒ¤Á 2014 19:57 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 3 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ FranceZii Siriseth
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  #3  
Old 26 µØÅÒ¤Á 2014, 20:00
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Old 26 µØÅÒ¤Á 2014, 20:41
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Old 26 µØÅÒ¤Á 2014, 21:15
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Old 26 µØÅÒ¤Á 2014, 22:27
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¼ÁÅͧãªé¨Ó¹Ç¹àªÔ§«é͹·´æ´ÙáÅéÇÂѧäÁèËÅØ´¤ÃѺ

ÍÕ¡ÇÔ¸ÕÃÒ¡ÊÁ¡Òà äÁèÃÙé¼Á仨ÓÊѺ¡Ñºâ¨·ÂìÊÁÒ¤ÁËÃ×Íà»ÅèÒ

¤ÃèÒÇæÅͧÊÃéÒ§ÊÁ¡Ò÷ÕèÁÕÊèǹ¡ÅѺ¢Í§ sin20,sin40,sin80 à»ç¹ÃÒ¡´Ù
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Old 30 µØÅÒ¤Á 2014, 00:05
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Old 15 ¾ÄȨԡÒ¹ 2014, 20:50
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¨§ËÒ¤èҢͧ
$$(\frac{1}{sin(20)})^2+(\frac{1}{sin(40)})^2+(\frac{1}{sin(80) })^2 $$

»Å.ÁØÁà»ç¹Í§ÈÒ
$=\frac{(sin 40sin 80)^2+(sin 20sin80)^2+(sin20sin40)^2}{(sin 20sin 40sin80)^2} $
$=\frac{(sin 40cos 10)^2+(sin 20cos 10)^2+(sin20sin40)^2}{(sin 20sin 40cos 10)^2}$
$=\frac{({\frac{sin 50+sin30}{2}})^2 +({\frac{sin 30+sin10}{2}})^2+({\frac{cos 20+cos 60}{2}})^2}{(\frac{\sqrt{3}}{8})^2}$
$=\frac{\frac{1}{4}[(cos 40+\frac{1}{2})^2 +(\frac{1}{2}+sin 10)^2+(cos 20-\frac{1}{2})^2]}{(\frac{\sqrt{3}}{8})^2}$
$=\frac{\frac{1}{4}[(cos^2 40+cos 40+\frac{1}{4})+(sin^2 10+sin 10+\frac{1}{4})+(cos^2 20-cos 20+\frac{1}{4})]}{(\frac{\sqrt{3}}{8})^2}$
$=\frac{\frac{1}{4}(cos^2 40+sin^2 10+cos^2 20+cos 40+sin 10-cos 20+\frac{3}{4})}{(\frac{\sqrt{3}}{8})^2}$
$=\frac{\frac{1}{4}(\frac{cos 80+1}{2} +\frac{1-cos 20}{2}+\frac{cos 40+1}{2}+cos 40+sin 10-cos 20+\frac{3}{4})}{(\frac{\sqrt{3}}{8})^2}$
$=\frac{\frac{1}{4}(\frac{sin 10+1}{2} +\frac{1-cos 20}{2}+\frac{cos 40+1}{2}+cos 40+sin 10-cos 20+\frac{3}{4})}{(\frac{\sqrt{3}}{8})^2}$
$=\frac{\frac{3}{8}(cos 40+sin 10-cos 20 +\frac{3}{2})}{(\frac{\sqrt{3}}{8})^2}$
$=\frac{\frac{3}{8}(0 +\frac{3}{2})}{\frac{3}{64}}$
$=12$
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