$\begin{array}{ll}\\ 1.&\textrm{Tunjukkan bahwa}\\ &\cot 7\displaystyle \frac{1}{2}^{0}=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}\\\\ &\textbf{Bukti}\\ &\begin{aligned}\cot \alpha &=\displaystyle \frac{\cos \alpha }{\sin \alpha }=\displaystyle \frac{2\cos ^{2}\alpha }{2\sin \alpha \cos \alpha }\\ &=\displaystyle \frac{1+\cos 2\alpha }{\sin 2\alpha }\\ &=\displaystyle \frac{1+\cos 15^{\circ} }{\sin 15^{\circ} }\\ \cot 7\displaystyle \frac{1}{2}^{0}&=\displaystyle \frac{1+\cos (45-30) }{\sin (45-30) }\\\\ &=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}\\ &=\displaystyle \frac{1+\displaystyle \frac{1}{4}\sqrt{6}+\frac{1}{4}\sqrt{2}}{\displaystyle \frac{1}{4}\sqrt{6}-\frac{1}{4}\sqrt{2}}\\ &=\displaystyle \frac{4+\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}\\ &=\displaystyle \frac{\left ( 4+\sqrt{6}+\sqrt{2} \right )\left ( \sqrt{6}+\sqrt{2} \right )}{6-2}\\ &=\displaystyle \frac{4\sqrt{6}+4\sqrt{2}+6+2\sqrt{3}+2\sqrt{3}+2}{4}\\ &=\displaystyle \frac{8+4\sqrt{2}+4\sqrt{3}+4\sqrt{6}}{4}\\ &=2+\sqrt{2}+\sqrt{3}+\sqrt{6}\\ &=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}\qquad \blacksquare \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 2.&\textrm{Tentukan nilai eksak dari}\: \: \sin 18^{\circ}\\\\ &\textbf{Jawab}\\ &\begin{aligned}&\textrm{Diketahui}\: \: 4(18^{\circ})=72^{\circ}=90^{\circ}-18^{\circ}\\ &\textrm{maka kita pilih}\: \: x=18^{\circ}.\: \textrm{Selanjutnya}\\ &\sin 4x=\sin (90^{\circ}-x)=\cos x\\ &\Leftrightarrow 2\sin 2x\cos 2x=\cos x\\ &\Leftrightarrow 2(2\sin x\cos x)(1-2\sin ^{2}x)=\cos x\\ &\Leftrightarrow 4\sin x(1-2\sin ^{2}x)=1\\ &\Leftrightarrow 4\sin x-8\sin ^{3}x-1=0\\ &\Leftrightarrow 8\sin ^{3}x-4\sin x+1=0\\ &\Leftrightarrow \left ( \sin x-1 \right )\left (4\sin ^{2}+2\sin x-1 \right )=0\\ &\Leftrightarrow 2\sin x=1\: \: \textrm{atau}\: \: 2\sin x=\displaystyle \frac{-1\pm \sqrt{5}}{2}\\ &\textrm{Nilai yang mungkin untuk}\: \: 2\sin x\: \: \textrm{dari}\\ &\textrm{ketiga nilai di atas adalah}\: \: \displaystyle \frac{-1+\sqrt{5}}{2}\\ &\textrm{Sehinga nilai dari}\\ &2\sin x=\displaystyle \frac{-1+\sqrt{5}}{2}\\ &\Leftrightarrow \sin x=\displaystyle \frac{-1+\sqrt{5}}{4}\\ &\textrm{karena}\: \: x=18^{\circ},\: \: \textrm{akan didapatkan}\\ &\textrm{nilai}\: \: \sin 18^{\circ}=\displaystyle \frac{-1+\sqrt{5}}{4}\\ &\textrm{Jadi, nilai eksak dari}\: \: \sin 18^{\circ}=\displaystyle \frac{\sqrt{5}-1}{4} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 3.&\textrm{Tunjukkan bahwa}\\ & \tan 11\displaystyle \frac{1}{4}^{\circ}=\sqrt{4+2\sqrt{2}}-\sqrt{2}-1\\\\ &\textbf{Bukti}\\ &\begin{aligned}&\textrm{Langkah awal}\\ &\tan 22\frac{1}{2}^{\circ}=\displaystyle \frac{\sin 22\displaystyle \frac{1}{2}^{\circ}}{\cos 22\displaystyle \frac{1}{2}^{\circ}}=\displaystyle \frac{2\sin 22\displaystyle \frac{1}{2}^{\circ}\sin 22\displaystyle \frac{1}{2}^{\circ}}{2\sin 22\displaystyle \frac{1}{2}^{\circ}\cos 22\displaystyle \frac{1}{2}^{\circ}}\\ &\qquad =\displaystyle \frac{1-\cos 45^{\circ}}{\sin 45^{\circ}}=\displaystyle \frac{1-\displaystyle \frac{1}{2}\sqrt{2}}{\displaystyle \frac{1}{2}\sqrt{2}}=\displaystyle \frac{2-\sqrt{2}}{\sqrt{2}}\\ &\qquad =\displaystyle \frac{2-\sqrt{2}}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=\displaystyle \frac{2\sqrt{2}-2}{2}\\ &\qquad =\sqrt{2}-1\\ &\textrm{Langkah berikutnya}\\ &\textrm{Misalkan}\: \: \tan 11\frac{1}{4}^{\circ}=x,\: \: \textrm{maka}\\ &\tan 22\displaystyle \frac{1}{2}^{\circ}=\displaystyle \frac{2\tan 11\displaystyle \frac{1}{4}^{\circ}}{1-\tan ^{2}11\displaystyle \frac{1}{4}^{\circ}}\\ &\Leftrightarrow \sqrt{2}-1=\displaystyle \frac{2x}{1-x^{2}}\\ &\Leftrightarrow \displaystyle \frac{1}{\sqrt{2}-1}=\displaystyle \frac{1-x^{2}}{2x}\\ &\Leftrightarrow x^{2}+2\left ( \sqrt{2}+1 \right )x-1=0\\ &\Leftrightarrow \: x_{_{1,2}}=\displaystyle \frac{-2\left ( \sqrt{2}+1 \right )\pm \sqrt{4\left ( \sqrt{2}+1 \right )^{2}+4}}{2}\\ &\Leftrightarrow \: x_{_{1,2}}= -\left ( \sqrt{2}+1 \right )\pm \sqrt{3+2\sqrt{2}+1}\\ &\Leftrightarrow \: x_{_{1,2}}=-\sqrt{2}-1\pm \sqrt{4+2\sqrt{2}}\\ &\textrm{Ambil yang nilai positif saja}\\ &\textrm{Sehingga}\\ & \tan 11\displaystyle \frac{1}{4}^{\circ}=-\sqrt{2}-1+\sqrt{4+2\sqrt{2}}\quad \blacksquare \\ \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 4.&\textrm{Jika}\: \: \alpha ,\: \beta ,\: \: \textrm{dan}\: \: \gamma \: \: \textrm{adalah sudut}\\ & \textrm{pada segitiga ABC, buktikan bahwa}\\ &\tan \displaystyle \frac{\alpha }{2}\tan \displaystyle \frac{\beta }{2}+\tan \displaystyle \frac{\beta }{2}\tan \displaystyle \frac{\gamma }{2}+\tan \displaystyle \frac{\gamma }{2}\tan \displaystyle \frac{\alpha }{2}=1\\\\ &\textbf{Bukti}\\ &\begin{aligned}&\textrm{Dikatahui bahwa}\\ &\alpha +\beta =180^{\circ}-\gamma \\ & \textrm{atau}\: \: \displaystyle \frac{1}{2}(\alpha +\beta )=\displaystyle \frac{1}{2}\left ( 180^{\circ}-\gamma \right )=90^{\circ}-\displaystyle \frac{\gamma }{2}\\ &\textrm{Maka}\\ &\tan \left ( \displaystyle \frac{\alpha }{2}+\frac{\beta }{2} \right )=\tan \left ( 90^{\circ}-\displaystyle \frac{\gamma }{2} \right )=\cot \displaystyle \frac{\gamma }{2}\\ &\Leftrightarrow \tan \left ( \displaystyle \frac{\alpha }{2}+\frac{\beta }{2} \right )=\displaystyle \frac{1}{\tan \displaystyle \frac{\gamma }{2}}\\ &\Leftrightarrow \displaystyle \frac{\tan \displaystyle \frac{\alpha }{2}+\tan \displaystyle \frac{\beta }{2}}{1-\tan \displaystyle \frac{\alpha }{2}\tan \displaystyle \frac{\beta }{2}}=\displaystyle \frac{1}{\tan \displaystyle \frac{\gamma }{2}}\\ &\Leftrightarrow \tan \displaystyle \frac{\alpha }{2}\tan \displaystyle \frac{\gamma }{2}+\tan \displaystyle \frac{\beta }{2}\tan \displaystyle \frac{\gamma }{2}=1-\tan \displaystyle \frac{\alpha }{2}\tan \displaystyle \frac{\beta }{2}\\ &\Leftrightarrow \tan \displaystyle \frac{\alpha }{2}\tan \displaystyle \frac{\beta }{2}+\tan \displaystyle \frac{\beta }{2}\tan \displaystyle \frac{\gamma }{2}+\tan \displaystyle \frac{\gamma }{2}\tan \displaystyle \frac{\alpha }{2}=1\quad \blacksquare \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 5.&\textrm{Jika}\: \: \alpha ,\: \beta ,\: \: \textrm{dan}\: \: \gamma \: \: \textrm{adalah sudut}\\ & \textrm{pada segitiga ABC, buktikan bahwa}\\ &\cos \alpha +\cos \beta +\cos \gamma =1+4\sin \displaystyle \frac{\alpha }{2}\sin \displaystyle \frac{\beta }{2}\sin \displaystyle \frac{\gamma }{2}\\\\ &\textbf{Bukti}\\ &\begin{aligned}&\textrm{Dikatahui bahwa}\\ &\alpha +\beta =180^{\circ}-\gamma \\ & \textrm{atau}\: \: \displaystyle \frac{1}{2}(\alpha +\beta )=\displaystyle \frac{1}{2}\left ( 180^{\circ}-\gamma \right )=90^{\circ}-\displaystyle \frac{\gamma }{2}\\ &\textrm{Maka}\\ &\cos \alpha +\cos \beta +\cos \gamma =2\cos \left ( \displaystyle \frac{\alpha +\beta }{2} \right )\cos \left ( \displaystyle \frac{\alpha -\beta }{2} \right )\\ &\qquad\qquad\qquad\qquad\qquad +\cos \left ( 180^{\circ}-\left ( \alpha +\beta \right ) \right )\\ &\Leftrightarrow \cos \alpha +\cos \beta +\cos \gamma =2\cos \left ( \displaystyle \frac{\alpha +\beta }{2} \right )\cos \left ( \displaystyle \frac{\alpha -\beta }{2} \right )\\ &\qquad\qquad\qquad\qquad\qquad -\cos (\alpha +\beta )\\ &\Leftrightarrow \cos \alpha +\cos \beta +\cos \gamma =2\cos \left ( \displaystyle \frac{\alpha +\beta }{2} \right )\cos \left ( \displaystyle \frac{\alpha -\beta }{2} \right )\\ &\qquad\qquad\qquad\qquad\qquad -2\cos ^{2}\left ( \displaystyle \frac{\alpha +\beta }{2} \right )+1\\ &\Leftrightarrow \cos \alpha +\cos \beta +\cos \gamma \\ &\quad\quad =2\cos \left ( \displaystyle \frac{\alpha +\beta }{2} \right )\left (\cos \left ( \displaystyle \frac{\alpha -\beta }{2} \right )-\cos \left ( \displaystyle \frac{\alpha +\beta }{2} \right ) \right )+1\\ &\Leftrightarrow \cos \alpha +\cos \beta +\cos \gamma \\ &\quad\quad =2\cos \left ( \displaystyle \frac{\alpha +\beta }{2} \right )\left ( -2\sin \displaystyle \frac{\alpha }{2}\sin \displaystyle \frac{-\beta }{2} \right )+1\\ &\Leftrightarrow \cos \alpha +\cos \beta +\cos \gamma \\ &\quad\quad =2\cos \left ( \displaystyle \frac{\alpha +\beta }{2} \right )\left ( 2\sin \displaystyle \frac{\alpha }{2}\sin \displaystyle \frac{\beta }{2} \right )+1\\ &\Leftrightarrow \cos \alpha +\cos \beta +\cos \gamma =2\cos \left ( \displaystyle \frac{\alpha +\beta }{2} \right )\left ( 2\sin \displaystyle \frac{\alpha }{2}\sin \displaystyle \frac{\beta }{2} \right )+1\\ &\Leftrightarrow \cos \alpha +\cos \beta +\cos \gamma =2\cos \left ( 90^{\circ}-\displaystyle \frac{\gamma }{2} \right )\left ( 2\sin \displaystyle \frac{\alpha }{2}\sin \displaystyle \frac{\beta }{2} \right )+1\\ &\Leftrightarrow \cos \alpha +\cos \beta +\cos \gamma =4\sin \displaystyle \frac{\gamma }{2}\left ( \sin \displaystyle \frac{\alpha }{2}\sin \displaystyle \frac{\beta }{2} \right )+1\\ &\Leftrightarrow \cos \alpha +\cos \beta +\cos \gamma =1+4\sin \displaystyle \frac{\alpha }{2}\sin \displaystyle \frac{\beta }{2}\sin \displaystyle \frac{\gamma }{2}\qquad \blacksquare \end{aligned} \end{array}$
DAFTAR PUSTAKA
- Bambang, S. 2012. Materi, Soal dan Penyelesaian Olimpiade Matematika Tingkat SMA/MA. Jakarta: BINA PRESTASI INSANI
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