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µÃÕ⡳ÁÔµÔ
ã¹ÊÒÁàËÅÕèÂÁABCÁÕÁØÁAà»ç¹ÁØÁ©Ò¡ sin B cos C+(sin C)( cos B)ÁÕ¤èÒà·èҡѺ¤èÒã¹¢éÍã´
CHOICE 1) $sin^2 B + cos^2 C$ 2) $sec^2 B+ tan^2 B$ 3) $cosec^2 B- cot^2 B$ 4) $2 cos B sin B$
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30 ¡Ñ¹ÂÒ¹ 2009 15:57 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ jspan |
#2
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the answer is 3)
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site:mathcenter.net ¤Ó¤é¹ |
#3
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co-function of trigon..sin(A)=cos(90-A)
áÅСéÍä»àÃ×èÍÂæ¤ÃѺ
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My stAtUs ·ÓäÁÂÔè§àÃÕ¹ áÅéÇÂÔè§â§èËÇèÒÒ 29 ¡Ñ¹ÂÒ¹ 2009 16:20 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ ~king duk kong~ |
#4
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ÍéÒ§ÍÔ§:
ÂѧäÁèà¤ÂàÃÕ¹µÃÕ⡹´éÇ«éÓ ÍÔÍÔ
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#5
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B+C=90
B=90-C ´Ñ§¹Ñé¹ sinB=cosC,sinC=cosB ¹èÒ¨Ðä»ä´éáÅéǹФÃѺ
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My stAtUs ·ÓäÁÂÔè§àÃÕ¹ áÅéÇÂÔè§â§èËÇèÒÒ |
#6
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sinB=cosC
sinC=cosB $sin B cos C+(sin C)( cos B) = sin^2B+cos^2B = 1 (¨Ò¡àÍ¡Åѡɳì)$ $ªéÍ¢éÍ3) cosec^2B-cot^2B = (1+cot^2B)-cot^2B (¨Ò¡àÍ¡Åѡɳì) = 1$ |
#7
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µÍº 3) ªÑÇÃìæ
ÇԸշӢͧàÃÒ¡ç¤×Í ÇÒ´ÃÙ» 3 àËÅÕèÂÁÁÒ 1 ÃÙ»ÍèйРáÅéÇ¡ç¨Ðä´éà»ç¹ÃÙ»´Ñ§¹Õé:- ¨Ò¡¹Ñ鹨Ðä´é sin B = $\frac{c}{d}$ cos C = $\frac{c}{d}$ sin C = $\frac{a}{b}$ cos B = $\frac{a}{b}$ $\therefore$ sinB = cosC áÅÐ sinC = cosB áÅéÇ¡ç¤Ô´áºº"¤Ø³ÍÂÒ¡à¡è§àÅ¢¤ÃѺ"ÍèÒ¤èСç¨Ðä´éà»ç¹¢éÍ 3) Ans
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#8
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µéͧãªéÊٵõÃÕ⡳Á.5ÍФÃѺ
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#9
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¤§äÁè¶Ö§¢¹Ò´¹Ñé¹ÁÑ駤ÃѺ à¾ÃÒÐ ã¹ Á.3 ¡çÁÕàÃÕ¹ Co-Function áÅéÇ¡çàÍ¡Åѡɳì 3 Ẻ ¤§à¾Õ§¾ÍµèÍ
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#10
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¤ÃѺ¼ÁÍèҹ⨷Âì¼Ô´ÍФÃѺ
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