Answer: C
\(\cos 15^{\circ} = \cos (45^{\circ}\:-\:30^{\circ})\)
\(\cos 15^{\circ} = \cos 45^{\circ} \cos 30^{\circ} + \sin 45^{\circ} \sin 30^{\circ}\)
\(\cos 15^{\circ} = \dfrac{1}{2}\sqrt{2}\cdot \dfrac{1}{2}\sqrt{3} + \dfrac{1}{2}\sqrt{2}\cdot \dfrac{1}{2}\)
\(\cos 15^{\circ} = \dfrac{1}{4}\sqrt{6} + \dfrac{1}{4}\sqrt{2}\)
\(\cos 15^{\circ} = \dfrac{1}{4}(\sqrt{6} + \sqrt{2})\)
Answer: B
\(\tan 105^{\circ} = \dfrac{\tan 60^{\circ} + \tan 45^{\circ}}{1\:-\:\tan 60^{\circ}\tan 45^{\circ}}\)
\(\tan 105^{\circ} = \dfrac{\sqrt{3} + 1}{1\:-\:\sqrt{3}\cdot 1}\)
\(\tan 105^{\circ} = \dfrac{\sqrt{3} + 1}{1\:-\:\sqrt{3}}\times \color{red}\dfrac{1 + \sqrt{3}}{1 + \sqrt{3}}\)
\(\tan 105^{\circ} = \dfrac{(\sqrt{3} + 1)(1 + \sqrt{3})}{1\:-\:3}\)
\(\tan 105^{\circ} = \dfrac{3 + 2\sqrt{3} + 1}{-2}\)
\(\tan 105^{\circ} = \dfrac{4 + 2\sqrt{3}}{-2}\)
\(\tan 105^{\circ} = -2\:-\:\sqrt{3}\)
Pada segitiga siku-siku ABC, diketahui bahwa \(\sin \text{A}\sin \text{B} = \dfrac{2}{5}\) dan \(\sin \text{(A − B)} = \dfrac{1}{5}a\). Nilai \(a = \dotso\)
Answer: C
\(\sin \text{(A − B)} = \dfrac{a}{5}\)
\(\cos \text{(A − B)} = \dfrac{\sqrt{5^2\:-\:a^2}}{5}\)
\(\cos \text{(A − B)} = \dfrac{\sqrt{5^2\:-\:a^2}}{5}\)
\(\cos \text{A}\cos \text{B} + \sin \text{A} \sin \text{B} = \dfrac{\sqrt{5^2\:-\:a^2}}{5}\)
\(\cos \text{A}\cos \text{B} + \dfrac{2}{5} = \dfrac{\sqrt{5^2\:-\:a^2}}{5}\)
\(\cos \text{A}\cos \text{B} = \dfrac{\sqrt{25\:-\:a^2}}{5}\:-\: \dfrac{2}{5}\)
\(\cos \text{A}\cos \text{B} = \dfrac{\sqrt{25\:-\:a^2}\:-\:2}{5}\:\:\:\:\:\color{blue}\text{persamaan 1}\)
Pada segitiga siku-siku ABC berlaku A + B + C = 180°, dengan C adalah sudut siku-siku
A + B + 90° = 180°
A + B = 180° − 90° = 90°
\(\cos (\text{A + B}) = \cos \text{A}\cos \text{B}\:-\:\sin \text{A} \sin \text{B}\)
\(\cos 90^{\circ} = \cos \text{A}\cos \text{B}\:-\:\dfrac{2}{5}\)
\(0 = \cos \text{A}\cos \text{B}\:-\:\dfrac{2}{5}\)
\(\cos \text{A}\cos \text{B}= \dfrac{2}{5}\:\:\:\:\:\color{blue}\text{persamaan 2}\)
Dengan menyamakan persamaan 1 dan 2, diperoleh:
\(\dfrac{\sqrt{25\:-\:a^2}\:-\:2}{\cancel{5}} = \dfrac{2}{\cancel{5}}\)
\(\sqrt{25\:-\:a^2}\:-\:2 = 2\)
\(\sqrt{25\:-\:a^2} = 4\:\:\:\:\:\color{blue}\text{kuadratkan kedua ruas}\)
\(25\:-\:a^2 = 16\)
\(a^2 = 25\:-\: 16 = 9\)
\(a = \pm \sqrt{9}\)
\(a = \pm 3\)
\(a = 3\)
Karena \(\sin \text{(A − B)} = \dfrac{1}{5}a\) dan A − B sudut di kuadran I, maka nilai \(a\) harus bernilai positif
Diketahui dalam segitiga lancip ABC, \(\sin \text{C} = \dfrac{2}{13}\sqrt{13}\) dan \(\tan \text{A}\cdot \tan \text{B} = 13\). Nilai \(\tan \text{A} + \tan \text{B} = \dotso\)
Answer: D
Dalam segitiga lancip ABC, berlaku A + B + C = 180°, sehingga C = 180° − (A + B)
Karena diketahui nilai \(\sin \text{C} = \dfrac{2}{13}\sqrt{13} = \dfrac{2}{\sqrt{13}}\), maka dengan bantuan segitiga siku-siku kita dapat menghitung nilai \(\tan \text{C}\)
\(\tan \text{C} = \dfrac{2}{3}\)
\(\tan [180° − (\text{A + B})] = \dfrac{2}{3}\)
\(-\tan (\text{A + B}) = \dfrac{2}{3}\)
\(\tan (\text{A + B}) = -\dfrac{2}{3}\)
\(\dfrac{\tan \text{A} + \tan \text{B}}{1\:-\:\tan \text{A}\tan \text{B}} = -\dfrac{2}{3}\)
\(\dfrac{\tan \text{A} + \tan \text{B}}{1\:-\:13} = -\dfrac{2}{3}\)
\(\dfrac{\tan \text{A} + \tan \text{B}}{-12} = -\dfrac{2}{3}\:\:\:\:\:\color{blue}\text{kali silang}\)
\(\tan \text{A} + \tan \text{B} = -\dfrac{2}{\cancel{3}} \times (-\cancelto{4}{12})\)
\(\tan \text{A} + \tan \text{B} = 8\)
Jika \(\cos (\text{A − B}) = \dfrac{3}{5}\) dan \(\cos (\text{A + B}) = \dfrac{2}{3}\), maka \(\tan \text{A} \cdot \tan \text{B} = \dotso\)
Answer: A
Diketahui \(\cos (\text{A − B}) = \dfrac{3}{5}\) dan \(\cos (\text{A + B}) = \dfrac{2}{3}\)
\(\cos \text{A}\cos \text{B} + \sin \text{A} \sin \text{B} = \dfrac{3}{5}\:\:\:\:\:\color{blue}\text{persamaan 1}\)
\(\cos \text{A}\cos \text{B} \:-\: \sin \text{A} \sin \text{B} = \dfrac{2}{3}\:\:\:\:\:\color{blue}\text{persamaan 2}\)
Dengan metode eliminasi,
Kurangi persamaan (1) dan (2), sehingga diperoleh:
\(2\sin \text{A} \sin \text{B} = \dfrac{3}{5}\:-\: \dfrac{2}{3}\)
\(2\sin \text{A} \sin \text{B} = \dfrac{9}{15}\:-\: \dfrac{10}{15}\)
\(2\sin \text{A} \sin \text{B} = -\dfrac{1}{15}\)
\(\sin \text{A} \sin \text{B} = -\dfrac{1}{30}\)
Tambahkan persamaan (1) dan (2), sehingga diperoleh:
\(2\cos \text{A}\cos \text{B} = \dfrac{3}{5}+\dfrac{2}{3}\)
\(2\cos \text{A}\cos \text{B} = \dfrac{9}{15}+\dfrac{10}{15}\)
\(2\cos \text{A}\cos \text{B} = \dfrac{19}{15}\)
\(\cos \text{A}\cos \text{B} = \dfrac{19}{30}\)
\(\tan \text{A} \cdot \tan \text{B} = \dfrac{\sin \text{A} \sin \text{B}}{\cos \text{A}\cos \text{B}}\)
\(\tan \text{A} \cdot \tan \text{B} = \dfrac{-\dfrac{1}{30}}{\dfrac{19}{30}}\)
\(\tan \text{A} \cdot \tan \text{B} = -\dfrac{1}{\cancel{30}} \times \dfrac{\cancel{30}}{19}\)
\(\tan \text{A} \cdot \tan \text{B} = -\dfrac{1}{19}\)
Jika \(\tan (x + y) = 1\) dan \(\tan (2x + y) = \sqrt{3}\), dengan \(x\) dan \(y\) sudut lancip, maka nilai \(\tan (3x + 2y) = \dotso\)
Answer: A
\(\tan (3x + 2y) = \tan [(x + y) + (2x + y)]\)
\(\tan (3x + 2y) = \dfrac{\tan (x + y) + \tan (2x + y)}{1\:-\:\tan (x + y) \tan (2x + y)}\)
\(\tan (3x + 2y) = \dfrac{1 + \sqrt{3}}{1\:-\:1\cdot \sqrt{3}}\)
\(\tan (3x + 2y) = \dfrac{1 + \sqrt{3}}{1\:-\:\sqrt{3}}\times \color{red}\dfrac{1 + \sqrt{3}}{1 + \sqrt{3}}\)
\(\tan (3x + 2y) = \dfrac{(1 + \sqrt{3})^2}{1\:-\:3}\)
\(\tan (3x + 2y) = \dfrac{1 + 2\sqrt{3} + 3}{-2}\)
\(\tan (3x + 2y) = \dfrac{4 + 2\sqrt{3}}{-2}\)
\(\tan (3x + 2y) = -2\:-\:\sqrt{3}\)
