Soal 1
Jika \(\tan 40^{\circ} = \text{k}\), maka\(\dfrac{2\sin 20^{\circ} \cos 20^{\circ}}{2\:-\:4\cos^2 20^{\circ}} = \dotso\)
(A) \(-\dfrac{1}{4}\text{k}\)
(B) \(-\dfrac{1}{2}\text{k}\)
(C) \(\dfrac{1}{2}\text{k}\)
(D) \(\dfrac{1}{4}\text{k}\)
(E) \(\dfrac{2}{5}\text{k}\)
Answer: B
\(\dfrac{2\sin 20^{\circ} \cos 20^{\circ}}{2\:-\:4\cos^2 20^{\circ}}\)
\(\dfrac{\sin 40^{\circ}}{2(1\:-\:2\cos^2 20^{\circ})}\)
\(\dfrac{\sin 40^{\circ}}{-2(2\cos^2 20^{\circ}\:-\:1)}\)
\(\dfrac{\sin 40^{\circ}}{-2\cos 40^{\circ}}\)
\(-\dfrac{1}{2}\cdot \dfrac{\sin 40^{\circ}}{\cos 40^{\circ}}\)
\(-\dfrac{1}{2}\cdot \tan 40^{\circ}\)
\(-\dfrac{1}{2}\text{k}\)
Soal 2
\(\dfrac{1+ \sin 2x + \cos 2x}{-1\:-\:\sin 2x + \cos 2x} = \dotso\)
(A) \(-2\cot x\)
(B) \(-\cot x\)
(C) \(\cot x\)
(D) \(2\cot x\)
(E) \(3\cot x\)
Answer: B
\(\dfrac{1+ \sin 2x + \cos 2x}{-1\:-\:\sin 2x + \cos 2x}\)
\(\color{blue}\text{ubah cos 2x untuk menghilangkan angka 1}\)
\(\dfrac{1+ \sin 2x + 2\cos^2 x\:-\:1}{-1\:-\:\sin 2x + 1 \:-\:2\sin^2 x}\)
\(\dfrac{\sin 2x + 2\cos^2 x}{-\sin 2x \:-\: 2\sin^2 x}\)
\(\color{blue}\text{ubah sin 2x = 2 sin x cos x}\)
\(\dfrac{2\sin x \cos x + 2\cos^2 x}{-2\sin x \cos x \:-\: 2\sin^2 x}\:\:\:\:\:\color{blue}\text{faktorkan}\)
\(\dfrac{\cancel{2}\cos x \cancel{(\sin x + \cos x)}}{-\cancel{2}\sin x \cancel{(\cos x + \sin x)}}\)
\(-\dfrac{\cos x}{\sin x}\)
\(-\cot x\)
Soal 3
\(\cos 34^{\circ} + \sin 34^{\circ}\cdot \tan 17^{\circ} = \dotso\)
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Answer: A
\(\cos 34^{\circ} + \sin 34^{\circ}\cdot \tan 17^{\circ}\)
\(\cos 34^{\circ} + 2\sin 17^{\circ} \cancel{\cos 17^{\circ}}\cdot \dfrac{\sin 17^{\circ}}{\cancel{\cos 17^{\circ}}}\)
\(\color{purple}\cos 34^{\circ}\color{black} + 2\sin 17^{\circ} \cdot \sin 17^{\circ}\)
\(\color{purple}1\:-\:2\sin^2 17^{\circ} \color{black}+ 2\sin^2 17^{\circ}\)
\(1\)
Soal 4
\(2\sin 18^{\circ} \cos 36^{\circ} = \dotso\)
(A) \(\dfrac{1}{16}\)
(B) \(\dfrac{1}{12}\)
(C) \(\dfrac{1}{8}\)
(D) \(\dfrac{1}{4}\)
(E) \(\dfrac{1}{2}\)
Answer: E
Langkah 1: Menghitung \(\sin 18^{\circ}\)
\(\text{Misal}, \theta = 18^{\circ}\)
\(5\theta = 90^{\circ}\)
\(2\theta + 3\theta = 90^{\circ}\)
\(2\theta = 90^{\circ}\:-\:3\theta\)
\(\sin 2\theta = \sin (90^{\circ}\:-\:3\theta)\)
\(\sin 2\theta = \cos 3 \theta\)
\(2\sin \theta \cos \theta = 4 \cos^3 \theta \:-\: 3\cos \theta\)
\(2\sin \theta \cos \theta\:-\: 4 \cos^3 \theta + 3\cos \theta = 0\:\:\:\:\:\color{blue}\text{faktorkan}\)
\(\cos \theta(2\sin \theta \:-\: 4\cos^2 \theta + 3) = 0\)
\(\cos \theta [2\sin \theta \:-\: 4(1\:-\:\sin^2 \theta) + 3] = 0\)
\(\cos \theta (4\sin^2 \theta + 2\sin \theta\:-\: 1) = 0\)
\(4\sin^2 \theta + 2\sin \theta\:-\: 1 = 0\)
\(\sin \theta = \dfrac{-1 \pm \sqrt{2^2 \:-\:4(4)(-1)}}{2(4)} = \dfrac{-2 \pm \sqrt{20}}{8}\)
\(\sin 18^{\circ} = \dfrac{\sqrt{5}\:-\:1}{4}\)
Langkah 2: Menghitung \(\cos 36^{\circ}\)
\(\cos 36^{\circ} = 1\:-\:2\sin^2 18^{\circ}\)
\(\cos 36^{\circ} = 1\:-\:2\left(\dfrac{\sqrt{5}\:-\:1}{4}\right)^2\)
\(\cos 36^{\circ} = 1\:-\:2\left(\dfrac{5\:-\:2\sqrt{5} + 1}{16}\right)\)
\(\cos 36^{\circ} = 1\:-\:\left(\dfrac{6\:-\:2\sqrt{5}}{8}\right)\)
\(\cos 36^{\circ} = \dfrac{4}{4}\:-\:\left(\dfrac{3\:-\:\sqrt{5}}{4}\right)\)
\(\cos 36^{\circ} = \dfrac{\sqrt{5} + 1}{4}\)
Langkah 3: Menghitung \(2\sin 18^{\circ} \cos 36^{\circ}\)
\(2\sin 18^{\circ} \cos 36^{\circ} = 2\left(\dfrac{\sqrt{5}\:-\:1}{4}\right)\left(\dfrac{\sqrt{5} + 1}{4}\right)\)
\(2\sin 18^{\circ} \cos 36^{\circ} = 2\times \dfrac{1}{4}\)
\(2\sin 18^{\circ} \cos 36^{\circ} = \dfrac{1}{2}\)