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#2
06 ÁԶعÒ¹ 2013, 21:11
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3) ãªéÊٵà $tan2\theta = \frac{2tan\theta}{1-tan^2\theta } $ -> ÇÒ´ÊÒÁàËÅÕèÂÁ¢Í§ÁØÁ $2\theta $ áÅéÇãªéÊٵà $sin4\theta = 2sin2\theta cos2\theta $ -> ä´é¤ÓµÍºáÅéǤÃѺ 4) ¡ÃШÒÂÊٵøÃÃÁ´Ò¡ç¹èÒ¨Ðä´é¹Ð¤ÃѺ 5) ¤Ù³´éÇ $ \frac{a}{cos\theta}$ ·Ñé§ÊÁ¡Òà ¨Ðä´é $\frac{sin\theta}{cos\theta }=tan\theta$ áÅéÇ¡çãªéÊٵà $tan2\theta $ àËÁ×͹¢éÍ 3-> ÇÒ´ÊÒÁàËÅÕèÂÁ 
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#4
06 ÁԶعÒ¹ 2013, 23:29
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$\frac{sin\theta }{a} = \frac{cos\theta }{b} $ $\frac{sin\theta }{cos\theta } = \frac{a}{b} $ $tan \theta = \frac{a}{b} $ $tan2\theta = \frac{2tan\theta }{1-tan^2\theta } $ $tan2\theta = \frac{\frac{2a}{b} }{1-\left(\frac{a}{b}\right)^2} $ $tan2\theta = \frac{\frac{2a}{b} }{\frac{b^2-a^2}{b^2}} $ $tan2\theta = \frac{2ab}{b^2-a^2}$ ÇÒ´ ÊÒÁàËÅÕèÂÁ  ¨Ò¡ÊÒÁàËÅÕèÂÁ¢éÒ§µé¹ ¨Ðä´é¤èҢͧ $sin2\theta $ áÅÐ $cos2\theta $ ¤ÃѺ
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#5
07 ÁԶعÒ¹ 2013, 10:59
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1.$2 cosec 2x =\sec x cosec x$
$2 cosec 2x =\dfrac{2}{\sin 2x} =\frac{1}{\sin x \cos x} =\sec x cosec x$ 2.$\sec 2x=\dfrac{\sec^2 x}{2-\sec^2 x} $ $\sec 2x=\dfrac{1}{\cos 2x} =\dfrac{1}{2\cos^2x-1} $ $=\dfrac{1}{\cos^2x(2-\sec^2 x)} $ $=\dfrac{\sec^2 x}{2-\sec^2 x} $
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