Untuk mengerjakan soal diatas, ada beberapa data yang kita perlukan, yaitu:
- $cos\ 2\theta=cos^{2}\theta-sin^{2}\theta$
- $cos\ 4\theta=cos^{2}2\theta-sin^{2}2\theta$
- $cos\ \theta=cos^{2}\frac{1}{2}\theta-sin^{2}\frac{1}{2}\theta$
- $sin^{2}\theta+cos^{2}\theta=1$
- $sin^{2}\frac{1}{2}\theta+cos^{2}\frac{1}{2}\theta=1$
- $sin\ 2\theta=2\ sin\ \theta\ cos\ \theta$
- $sin\ 4\theta=2\ sin\ 2\theta\ cos\ 2\theta$
- $sin\ \theta=2\ sin\ \frac{1}{2}\theta\ cos\ \frac{1}{2}\theta$
Data-data yang kita peroleh diatas kita substituskan ke soal,
$a=\frac{cos\ \theta}{1-sin\ \theta}$
$a=\frac{cos^{2}\frac{1}{2}\theta-sin^{2}\frac{1}{2}\theta}{\left ( sin^{2}\frac{1}{2}\theta+cos^{2}\frac{1}{2}\theta \right )\left ( 2\ sin\ \frac{1}{2}\theta\ cos\ \frac{1}{2}\theta \right )}$
$a=\frac{\left (cos\frac{1}{2}\theta-sin\frac{1}{2}\theta \right )\left ( cos\frac{1}{2}\theta+sin\frac{1}{2}\theta \right )}{\left (cos\frac{1}{2}\theta-sin\frac{1}{2}\theta \right )^{2}}$
$a=\frac{\left (cos\frac{1}{2}\theta+sin\frac{1}{2}\theta \right )}{\left (cos\frac{1}{2}\theta-sin\frac{1}{2}\theta \right )}$
$a\left (cos\frac{1}{2}\theta-sin\frac{1}{2}\theta \right )=\left (cos\frac{1}{2}\theta+sin\frac{1}{2}\theta \right )$
$a\ cos\frac{1}{2}\theta-a\ sin\frac{1}{2}\theta=cos\frac{1}{2}\theta+sin\frac{1}{2}\theta$
$a\ cos\frac{1}{2}\theta-cos\frac{1}{2}\theta=a\ sin\frac{1}{2}\theta+sin\frac{1}{2}\theta$
$cos\frac{1}{2}\theta\left (a\ -1 \right )=sin\frac{1}{2}\theta \left (a\ +1 \right )$
$\frac{\left (a\ -1 \right )}{\left (a\ +1 \right )}=\frac{sin\frac{1}{2}\theta}{cos\frac{1}{2}\theta}$
$\frac{\left (a\ -1 \right )}{\left (a\ +1 \right )}=tan\frac{1}{2}\theta$
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