Pembahasan Trigonometri Rangkap UM UGM 2006 Matematika Ipa

Soal yang Akan Dibahas
Jika $ \frac{\cos \alpha }{1 - \sin \alpha } = a $ , untuk $ \alpha \neq \frac{\pi}{2}+2k\pi $ , maka $ \tan \frac{\alpha}{2} = .... $
A). $ \frac{a}{a+1} \, $ B). $ \frac{1}{a+1} \, $
C). $ \frac{a-1}{a+1} \, $ D). $ \frac{a+1}{a-1} \, $
E). $ \frac{a}{a-1} $

$\spadesuit $ Konsep Dasar
*). Rumus Sudut Rangkap :
$ \begin{align} \cos \alpha & = \cos ^2 \frac{\alpha}{2} - \sin ^2 \frac{\alpha}{2} \\ & = (\cos \frac{\alpha}{2} + \sin \frac{\alpha}{2})(\cos \frac{\alpha}{2} - \sin \frac{\alpha}{2}) \end{align} $
$ \sin \alpha = 2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2} $
*). Rumus identitas trigonometri :
$ \sin ^2 \frac{\alpha}{2} + \cos ^2 \frac{\alpha}{2} = 1 $
*). Rumus perbandingan sederhana :
$ \tan \frac{\alpha}{2} = \frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}} $
*). Bentuk pemfaktoran : $ a^2 + b^2 - 2ab = (a-b)^2 $

$\clubsuit $ Pembahasan
*). Menyelesaikan Soal :
$ \begin{align} \frac{\cos \alpha }{1 - \sin \alpha } & = a \\ \frac{(\cos \frac{\alpha}{2} + \sin \frac{\alpha}{2})(\cos \frac{\alpha}{2} - \sin \frac{\alpha}{2}) }{1 - 2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2} } & = a \\ \frac{(\cos \frac{\alpha}{2} + \sin \frac{\alpha}{2})(\cos \frac{\alpha}{2} - \sin \frac{\alpha}{2}) }{ \sin ^2 \frac{\alpha}{2} + \cos ^2 \frac{\alpha}{2} - 2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2} } & = a \\ \frac{(\cos \frac{\alpha}{2} + \sin \frac{\alpha}{2})(\cos \frac{\alpha}{2} - \sin \frac{\alpha}{2}) }{ ( \cos \frac{\alpha}{2} - \sin \frac{\alpha}{2})^2 } & = a \\ \frac{(\cos \frac{\alpha}{2} + \sin \frac{\alpha}{2})(\cos \frac{\alpha}{2} - \sin \frac{\alpha}{2}) }{ ( \cos \frac{\alpha}{2} - \sin \frac{\alpha}{2})( \cos \frac{\alpha}{2} - \sin \frac{\alpha}{2}) } & = a \\ \frac{\cos \frac{\alpha}{2} + \sin \frac{\alpha}{2} }{ \cos \frac{\alpha}{2} - \sin \frac{\alpha}{2} } & = a \\ \cos \frac{\alpha}{2} + \sin \frac{\alpha}{2} & = a( \cos \frac{\alpha}{2} - \sin \frac{\alpha}{2} ) \\ \cos \frac{\alpha}{2} + \sin \frac{\alpha}{2} & = a \cos \frac{\alpha}{2} - a \sin \frac{\alpha}{2} \\ a \sin \frac{\alpha}{2} + \sin \frac{\alpha}{2} & = a \cos \frac{\alpha}{2} - \cos \frac{\alpha}{2} \\ (a + 1) \sin \frac{\alpha}{2} & = (a - 1 ) \cos \frac{\alpha}{2} \\ \frac{ \sin \frac{\alpha}{2} }{\cos \frac{\alpha}{2}} & = \frac{(a - 1 )}{(a+1)} \\ \tan \frac{\alpha}{2} & = \frac{a - 1 }{a+1} \end{align} $
Jadi, nilai $ \tan \frac{\alpha}{2} = \frac{a - 1 }{a+1} . \, \heartsuit $

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