PAS MATEMATIKA BLOG: Contoh Soal 1 (Segitiga dan Trigonometri)

$\begin{array}{ll} 1. & Tunjukkan bahwa \\ \cot 7 {\frac{1}{2}}^{0} = \sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{6} \\ Bukti \\ \begin{aligned} \cot α & = \frac{\cos α}{\sin α} = \frac{2 \cos^{2} α}{2 \sin α \cos α} \\ = \frac{1 + \cos 2 α}{\sin 2 α} \\ = \frac{1 + \cos 15^{\circ}}{\sin 15^{\circ}} \\ \cot 7 {\frac{1}{2}}^{0} & = \frac{1 + \cos (45 - 30)}{\sin (45 - 30)} \\ = \sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{6} \\ = \frac{1 + \frac{1}{4} \sqrt{6} + \frac{1}{4} \sqrt{2}}{\frac{1}{4} \sqrt{6} - \frac{1}{4} \sqrt{2}} \\ = \frac{4 + \sqrt{6} + \sqrt{2}}{\sqrt{6} - \sqrt{2}} \\ = \frac{(4 + \sqrt{6} + \sqrt{2}) (\sqrt{6} + \sqrt{2})}{6 - 2} \\ = \frac{4 \sqrt{6} + 4 \sqrt{2} + 6 + 2 \sqrt{3} + 2 \sqrt{3} + 2}{4} \\ = \frac{8 + 4 \sqrt{2} + 4 \sqrt{3} + 4 \sqrt{6}}{4} \\ = 2 + \sqrt{2} + \sqrt{3} + \sqrt{6} \\ = \sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{6} ◼ \end{aligned} \end{array}$ .

$\begin{array}{ll} 2. & Tentukan nilai eksak dari \sin 18^{\circ} \\ Jawab \\ \begin{aligned} Diketahui 4 (18^{\circ}) = 72^{\circ} = 90^{\circ} - 18^{\circ} \\ maka kita pilih x = 18^{\circ} . Selanjutnya \\ \sin 4 x = \sin (90^{\circ} - x) = \cos x \\ \Leftrightarrow 2 \sin 2 x \cos 2 x = \cos x \\ \Leftrightarrow 2 (2 \sin x \cos x) (1 - 2 \sin^{2} x) = \cos x \\ \Leftrightarrow 4 \sin x (1 - 2 \sin^{2} x) = 1 \\ \Leftrightarrow 4 \sin x - 8 \sin^{3} x - 1 = 0 \\ \Leftrightarrow 8 \sin^{3} x - 4 \sin x + 1 = 0 \\ \Leftrightarrow (\sin x - 1) (4 \sin^{2} + 2 \sin x - 1) = 0 \\ \Leftrightarrow 2 \sin x = 1 atau 2 \sin x = \frac{- 1 \pm \sqrt{5}}{2} \\ Nilai yang mungkin untuk 2 \sin x dari \\ ketiga nilai di atas adalah \frac{- 1 + \sqrt{5}}{2} \\ Sehinga nilai dari \\ 2 \sin x = \frac{- 1 + \sqrt{5}}{2} \\ \Leftrightarrow \sin x = \frac{- 1 + \sqrt{5}}{4} \\ karena x = 18^{\circ}, akan didapatkan \\ nilai \sin 18^{\circ} = \frac{- 1 + \sqrt{5}}{4} \\ Jadi, nilai eksak dari \sin 18^{\circ} = \frac{\sqrt{5} - 1}{4} \end{aligned} \end{array}$ .

$\begin{array}{ll} 3. & Tunjukkan bahwa \\ \tan 11 {\frac{1}{4}}^{\circ} = \sqrt{4 + 2 \sqrt{2}} - \sqrt{2} - 1 \\ Bukti \\ \begin{aligned} Langkah awal \\ \tan 22 {\frac{1}{2}}^{\circ} = \frac{\sin 22 {\frac{1}{2}}^{\circ}}{\cos 22 {\frac{1}{2}}^{\circ}} = \frac{2 \sin 22 {\frac{1}{2}}^{\circ} \sin 22 {\frac{1}{2}}^{\circ}}{2 \sin 22 {\frac{1}{2}}^{\circ} \cos 22 {\frac{1}{2}}^{\circ}} \\ = \frac{1 - \cos 45^{\circ}}{\sin 45^{\circ}} = \frac{1 - \frac{1}{2} \sqrt{2}}{\frac{1}{2} \sqrt{2}} = \frac{2 - \sqrt{2}}{\sqrt{2}} \\ = \frac{2 - \sqrt{2}}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{2 \sqrt{2} - 2}{2} \\ = \sqrt{2} - 1 \\ Langkah berikutnya \\ Misalkan \tan 11 {\frac{1}{4}}^{\circ} = x, maka \\ \tan 22 {\frac{1}{2}}^{\circ} = \frac{2 \tan 11 {\frac{1}{4}}^{\circ}}{1 - \tan^{2} 11 {\frac{1}{4}}^{\circ}} \\ \Leftrightarrow \sqrt{2} - 1 = \frac{2 x}{1 - x^{2}} \\ \Leftrightarrow \frac{1}{\sqrt{2} - 1} = \frac{1 - x^{2}}{2 x} \\ \Leftrightarrow x^{2} + 2 (\sqrt{2} + 1) x - 1 = 0 \\ \Leftrightarrow x_{_{1, 2}} = \frac{- 2 (\sqrt{2} + 1) \pm \sqrt{4 {(\sqrt{2} + 1)}^{2} + 4}}{2} \\ \Leftrightarrow x_{_{1, 2}} = - (\sqrt{2} + 1) \pm \sqrt{3 + 2 \sqrt{2} + 1} \\ \Leftrightarrow x_{_{1, 2}} = - \sqrt{2} - 1 \pm \sqrt{4 + 2 \sqrt{2}} \\ Ambil yang nilai positif saja \\ Sehingga \\ \tan 11 {\frac{1}{4}}^{\circ} = - \sqrt{2} - 1 + \sqrt{4 + 2 \sqrt{2}} ◼ \end{aligned} \end{array}$ .

$\begin{array}{ll} 4. & Jika α, β, dan γ adalah sudut \\ pada segitiga ABC, buktikan bahwa \\ \tan \frac{α}{2} \tan \frac{β}{2} + \tan \frac{β}{2} \tan \frac{γ}{2} + \tan \frac{γ}{2} \tan \frac{α}{2} = 1 \\ Bukti \\ \begin{aligned} Dikatahui bahwa \\ α + β = 180^{\circ} - γ \\ atau \frac{1}{2} (α + β) = \frac{1}{2} (180^{\circ} - γ) = 90^{\circ} - \frac{γ}{2} \\ Maka \\ \tan (\frac{α}{2} + \frac{β}{2}) = \tan (90^{\circ} - \frac{γ}{2}) = \cot \frac{γ}{2} \\ \Leftrightarrow \tan (\frac{α}{2} + \frac{β}{2}) = \frac{1}{\tan \frac{γ}{2}} \\ \Leftrightarrow \frac{\tan \frac{α}{2} + \tan \frac{β}{2}}{1 - \tan \frac{α}{2} \tan \frac{β}{2}} = \frac{1}{\tan \frac{γ}{2}} \\ \Leftrightarrow \tan \frac{α}{2} \tan \frac{γ}{2} + \tan \frac{β}{2} \tan \frac{γ}{2} = 1 - \tan \frac{α}{2} \tan \frac{β}{2} \\ \Leftrightarrow \tan \frac{α}{2} \tan \frac{β}{2} + \tan \frac{β}{2} \tan \frac{γ}{2} + \tan \frac{γ}{2} \tan \frac{α}{2} = 1 ◼ \end{aligned} \end{array}$ .

$\begin{array}{ll} 5. & Jika α, β, dan γ adalah sudut \\ pada segitiga ABC, buktikan bahwa \\ \cos α + \cos β + \cos γ = 1 + 4 \sin \frac{α}{2} \sin \frac{β}{2} \sin \frac{γ}{2} \\ Bukti \\ \begin{aligned} Dikatahui bahwa \\ α + β = 180^{\circ} - γ \\ atau \frac{1}{2} (α + β) = \frac{1}{2} (180^{\circ} - γ) = 90^{\circ} - \frac{γ}{2} \\ Maka \\ \cos α + \cos β + \cos γ = 2 \cos (\frac{α + β}{2}) \cos (\frac{α - β}{2}) \\ + \cos (180^{\circ} - (α + β)) \\ \Leftrightarrow \cos α + \cos β + \cos γ = 2 \cos (\frac{α + β}{2}) \cos (\frac{α - β}{2}) \\ - \cos (α + β) \\ \Leftrightarrow \cos α + \cos β + \cos γ = 2 \cos (\frac{α + β}{2}) \cos (\frac{α - β}{2}) \\ - 2 \cos^{2} (\frac{α + β}{2}) + 1 \\ \Leftrightarrow \cos α + \cos β + \cos γ \\ = 2 \cos (\frac{α + β}{2}) (\cos (\frac{α - β}{2}) - \cos (\frac{α + β}{2})) + 1 \\ \Leftrightarrow \cos α + \cos β + \cos γ \\ = 2 \cos (\frac{α + β}{2}) (- 2 \sin \frac{α}{2} \sin \frac{- β}{2}) + 1 \\ \Leftrightarrow \cos α + \cos β + \cos γ \\ = 2 \cos (\frac{α + β}{2}) (2 \sin \frac{α}{2} \sin \frac{β}{2}) + 1 \\ \Leftrightarrow \cos α + \cos β + \cos γ = 2 \cos (\frac{α + β}{2}) (2 \sin \frac{α}{2} \sin \frac{β}{2}) + 1 \\ \Leftrightarrow \cos α + \cos β + \cos γ = 2 \cos (90^{\circ} - \frac{γ}{2}) (2 \sin \frac{α}{2} \sin \frac{β}{2}) + 1 \\ \Leftrightarrow \cos α + \cos β + \cos γ = 4 \sin \frac{γ}{2} (\sin \frac{α}{2} \sin \frac{β}{2}) + 1 \\ \Leftrightarrow \cos α + \cos β + \cos γ = 1 + 4 \sin \frac{α}{2} \sin \frac{β}{2} \sin \frac{γ}{2} ◼ \end{aligned} \end{array}$

DAFTAR PUSTAKA

Bambang, S. 2012. Materi, Soal dan Penyelesaian Olimpiade Matematika Tingkat SMA/MA. Jakarta: BINA PRESTASI INSANI

PAS MATEMATIKA BLOG

Contoh Soal 1 (Segitiga dan Trigonometri)

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