#1
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µÃÕ⡳
21 àÁÉÒ¹ 2019 12:24 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ Hutchjang |
#2
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¢éͶÒÁà¾ÔèÁÍÕ¡¢éÍ à¡ÕèèÂǡѺµÃÕ⡳´éǤÃѺ |
#3
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ÍéÒ§ÍÔ§:
21 àÁÉÒ¹ 2019 17:15 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ Hutchjang |
#4
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ÍéÒ§ÍÔ§:
àËç¹ä´éªÑ´ÇèÒ cos x = sin y = cos z = 1 à·èÒ¹Ñé¹
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The Lost Emic <<-- ˹ѧÊ×Íà©Å¢éÍÊͺÃдѺ»ÃжÁ¹Ò¹ÒªÒµÔ EMIC ¤ÃÑ駷Õè 1 - ¤ÃÑ駷Õè 8 ªØ´ÊØ´·éÒ ËŧÁÒ 21 àÁÉÒ¹ 2019 19:17 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ gon |
#5
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ÍéÒ§ÍÔ§:
¢é͹Õé¨Ø´ËÅÑ¡ æ ¤×Í¡ÒÃá»Å§ $\cos$ ã¹à·ÍÁ¢Í§´éÒ¹ $a,b,c$ â´Âãªé¡®¢Í§ $\cos$ ¤ÃѺ Åͧ¤Ô´´Ù¡è͹ä´é¤ÃѺ ¶éÒ¤Ô´äÁèÍÍ¡¡ç¡´´Ùà©ÅÂä´é¤ÃѺ \begin{align*}\left(\frac{\cos{A}-\cos{B}}{a-b}\right)c+\cos{C}&=\left(\frac{\frac{b^2+c^2-a^2}{2bc}-\frac{a^2+c^2-b^2}{2ac}}{a-b}\right)c+\frac{a^2+b^2-c^2}{2ab}\\&=\frac{c^2-(a+b)^2}{2ab}+\frac{a^2+b^2-c^2}{2ab}\\&=-1\end{align*} |
#6
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ÍéÒ§ÍÔ§:
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#7
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¢Íº¤Ø³ ¤Ø³ÍÒ Gon ÁÒ¡¤ÃѺ
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#8
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ÍéÒ§ÍÔ§:
$\left(\dfrac{\cos A-\cos B}{a-b}\right)c+\cos C=\dfrac{(\cos A-\cos B)\sin C}{\sin A-\sin B}+\cos C=\dfrac{\sin(A+C)-\sin(B+C)}{\sin A-\sin B}=-1$ |
#9
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ÍéÒ§ÍÔ§:
25 àÁÉÒ¹ 2019 08:51 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ Hutchjang |
#10
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$\left(\dfrac{\cos A-\cos B}{a-b}\right)c+\cos C=\dfrac{(\cos A-\cos B)\sin C}{\sin A-\sin B}+\cos C$
$\left(\dfrac{\cos A-\cos B}{a-b}\right)c=\left(\dfrac{\cos A-\cos B}{\frac{a}{c} -\frac{b}{c} }\right)$ $\frac{\sin A}{a}=\frac{\sin B}{b} =\frac{\sin C}{c} $ $\frac{a}{c}=\frac{\sin A}{\sin B} ,\frac{b}{c}=\frac{\sin B}{\sin B} $ $\dfrac{(\cos A-\cos B)\sin C}{\sin A-\sin B}+\cos C=\dfrac{\left(\,\cos A \sin C+\sin A \cos C\right) -\left(\,\cos B\sin C+\sin B\cos C\right) }{\sin A-\sin B} $ $=\dfrac{\sin(A+C)-\sin(B+C)}{\sin A-\sin B}$ $A+B+C=180^o$ $\sin(A+C)-\sin(B+C) =\sin (180^o-B)-\sin (180^o-A)=\sin B-\sin A$
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#11
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ÍéÒ§ÍÔ§:
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