\(\dfrac{\sin \text{A} + \sin 5\text{A} + \sin 9\text{A}}{\cos \text{A} + \cos 5\text{A} + \cos 9\text{A}}=\dotso\)
Answer: E
\(\dfrac{\sin \text{A} + \sin 9\text{A} + \sin 5\text{A}}{\cos \text{A} + \cos 9\text{A}+ \cos 5\text{A}}\)
\(\dfrac{2\sin \dfrac{1}{2}(\text{A} + \text{9A})\cos \dfrac{1}{2}(\text{A} \:-\: \text{9A})+\sin \text{5A}}{2\cos \dfrac{1}{2}(\text{A} + \text{9A})\cos \dfrac{1}{2}(\text{A} \:-\: \text{9A})+\cos \text{5A}}\)
\(\dfrac{2\sin \text{5A}\cdot \cos \text{4A} + \sin \text{5A}}{2\cos \text{5A}\cdot \cos \text{4A} + \cos \text{5A}}\)
\(\dfrac{\cancel{2}\sin \text{5A}\cancel{(\cos \text{4A} + 1)}}{\cancel{2}\cos \text{5A}\cancel{(\cos \text{4A} + 1)}}\)
\(\tan \text{5A}\)
Jika \(\text{A, B, dan C}\) adalah sudut-sudut dalam sebuah segitiga, maka \(\sin \text{2A} + \sin \text{2B} + \sin \text{2C} = \dotso\)
Answer: D
Dalam sebuah segitiga A + B + C = 180°, sehingga C = 180° − (A + B)
\(\sin \text{2A} + \sin \text{2B} + \sin \text{2C}\)
\(\sin \text{2A} + \sin \text{2B} + \sin 2[180^{\circ} \:-\: (\text{A + B})]\)
\(\sin \text{2A} + \sin \text{2B} + \sin [360^{\circ} \:-\: 2(\text{A + B})]\)
\(\color{purple}\sin \text{2A} + \sin \text{2B} \color{black}\:-\: \sin 2(\text{A + B})\)
\(\color{purple}2\sin \dfrac{1}{2}(\text{2A + 2B})\cos \dfrac{1}{2}(\text{2A – 2B}) \color{black}\:-\: 2\sin (\text{A + B})\cos (\text{A + B})\)
\(2\sin (\text{A + B})\cos (\text{A}\:-\:\text{B})\:-\:2\sin(\text{A + B})\cos (\text{A + B})\:\:\:\:\:\color{blue}\text{faktorkan}\)
\(2\sin (\text{A + B})\color{purple}[\cos (\text{A}\:-\:\text{B})\:-\:\cos (\text{A + B})]\)
\(2\sin (180^{\circ}\:-\:\text{C})\color{purple}[-2\sin \dfrac{1}{2}(\text{A − B + A + B})\sin \dfrac{1}{2} (\text{A − B − A − B})]\)
\(2\sin (180^{\circ}\:-\:\text{C})[-2\sin \dfrac{1}{2} (\text{2A})\sin \dfrac{1}{2} (\text{−2B})]\)
\(2\sin \text{C}(2\sin \text{A} \sin \text{B})\)
\(4\sin \text{A} \cdot \sin \text{B} \cdot \sin \text{C}\)
\(\dfrac{\sin 5^{\circ} + \sin 20^{\circ} + \sin 35^{\circ}}{\cos 35^{\circ} + \cos 20^{\circ} + \cos 5^{\circ}}= \dotso\)
Jawaban: C
Gunakan:
\(\color{blue}\sin \text{A} + \sin \text{B} = 2\sin \frac{1}{2}(\text{A + B})\cos \frac{1}{2} (\text{A − B})\)
\(\color{blue}\cos\text{A} + \cos \text{B} = 2\cos \frac{1}{2}(\text{A + B})\cos \frac{1}{2} (\text{A − B})\)
\(\dfrac{\sin 5^{\circ} + \sin 35^{\circ} + \sin 20^{\circ}}{\cos 5^{\circ} + \cos 35^{\circ} + \cos 20^{\circ}}\)
\(\dfrac{2\sin \frac{1}{2}(40^{\circ})\cos \frac{1}{2}(-30^{\circ}) + \sin 20^{\circ}}{2\cos \frac{1}{2} (40^{\circ}) \cos \frac{1}{2} (-30^{\circ}) + \cos 20^{\circ}}\)
\(\dfrac{2\sin 20^{\circ} \cos 15^{\circ} + \sin 20^{\circ}}{2\cos 20^{\circ} \cos 15^{\circ} + \cos 20^{\circ} }\)
\(\dfrac{\sin 20^{\circ}\cancel{(2\cos 15^{\circ} + 1)}}{\cos 20^{\circ}\cancel{(2\cos 15^{\circ} + 1)}}\)
\(\dfrac{\sin 20^{\circ}}{\cos 20^{\circ}}\)
\(\tan 20^{\circ}\)
