อันนี้วิธีผมนะครับ มีวิธีไหนสั้นกว่านี้แบ่งปันด้วยครับ
จาก $sin(x)^2 = \frac{1-cos(2x)}{2}$
$$(\frac{1}{sin(20)})^2+(\frac{1}{sin(40})^2+(\frac{1}{sin(80) })^2 $$
$$ = \frac{2}{1-cos(40)}+\frac{2}{1-cos(80)}+\frac{2}{1-cos(160)}$$
$$ = 2*\frac{(1-cos(40))(1-cos(80))+(1-cos(40))(1-cos(160))+(1-cos(160))(1-cos(80))}{(1-cos(40))(1-cos(80)(1-cos(160)}$$
$$ =2*\frac{3-2(cos(40)+cos(80)+(cos(160))+cos(40)cos(80)+cos(40)cos(160)+cos(80)cos(160)}{1-(cos(40)+cos(80)+cos(160))+cos(40)cos(80)+cos(40)cos(160)+cos(80)cos(160)-cos(40)cos(80)cos(160)} $$
$cos(40)+cos(80)+cos(160)=0$
$$ =2*\frac{3+cos(40)cos(80)+cos(40)cos(160)+cos(80)cos(160)}{1+cos(40)cos(80)+cos(40)cos(160)+cos(80)cos(160)-cos(40)cos(80)cos(160)} $$
$cos(40)cos(80)+cos(40)cos(160)+cos(80)cos(160)=\frac{-3}{4}$
$cos(40)cos(80)cos(160)=\frac{-1}{8}$
จะได้ว่า
$$ =2*\frac{3+cos(40)cos(80)+cos(40)cos(160)+cos(80)cos(160)}{1+cos(40)cos(80)+cos(40)cos(160)+cos(80)cos(160)-cos(40)cos(80)cos(160)}$$
$$= 2*\frac{3-\frac{3}{4}}{1-\frac{3}{4}+\frac{1}{8}} $$
$$=12$$
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26 ตุลาคม 2014 19:57 : ข้อความนี้ถูกแก้ไขแล้ว 3 ครั้ง, ครั้งล่าสุดโดยคุณ FranceZii Siriseth
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