#1
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µÃÕ⡳ÁÔµÔ.
¢Íàà¹Ç¤Ô´Ë¹èͤÃѺ.
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#2
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¾Ô¨ÒÃ³Ò $f(x)=\sqrt{3}\sin x+\sqrt{5} \cos x$ ËÒªèǧ¢Í§ $f(x)$ $f'(x)=\sqrt{3}cos x-\sqrt{5}sin x=0$ $\tan x= \dfrac{\sqrt{3}}{\sqrt{5}}\Rightarrow -\sqrt{8}\leqslant \sin x \leqslant \sqrt{8}$ $\cos (15-y)+9=$¢éÒ§«éÒ $\leqslant 8$ áµè $-1\leqslant \cos \theta \leqslant 1$ ´Ñ§¹Ñé¹ ä´é $15-y = \pi$ $y=-165$ áÅéÇ¡éÍá·¹¤èÒ $\sin 75 =....$ àÍÒ¤ÃѺ Êèǹ $\sin x$ ËÒ¨Ò¡à§×è͹ä¢à·èҡѹ¢Í§ÍÊÁ¡Òà ä´é $\sin x= \dfrac{\sqrt{3}}{2\sqrt{2}}$ 08 ¾ÄÉÀÒ¤Á 2015 07:45 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ ¿Ô¹Ô¡«ìàËÔ¹¿éÒ |
#3
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ÍéÒ§ÍÔ§:
¢Íº¤Ø³¤ÃѺ¼Á.^^. |
#4
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¨Ò¡àÍ¡Åѡɳì $asinx+bcosx=ksin (x+\theta ) $ â´Â $k=\sqrt{a^2+b^2} $
ÊÁ¡ÒÃ⨷Âì¨Ðà»ç¹¨ÃÔ§àÁ×èÍ $k^2sin^2 (x+\theta )=8$ áÅÐ $cos (\frac {\pi}{12}-y)=-1$ ¹Ñ蹤×Í $sin(x+\theta )=\pm 1$ ´Ñ§¹Ñé¹ $x+\theta =\pm \frac {\pi}{2} $ áµè$sinx=sin(\frac {\pi}{2}-\theta )=cos\theta =\frac {\sqrt{3}}{\sqrt{8}}$ áÅÐ $ \frac {\pi}{12}-y=\pm \pi $ ¨Ðä´é $y=-\frac {11\pi}{12};\frac {13\pi}{12} $ $cosy=-cos\frac {\pi}{12}=-\frac{\sqrt{3}+1}{2\sqrt{2} }$ ¹Óä»á·¹ $\frac {sinx}{cosy}=-\frac{\sqrt{3}}{\sqrt{3}+1}=\frac{\sqrt{3}-3}{2} $ 09 ¾ÄÉÀÒ¤Á 2015 00:46 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ artty60 |
#5
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ÍéÒ§ÍÔ§:
¢Íº¤Ø³¤ÃѺ. ^^ . |
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