5. \( \operatorname{Tg} 30^{\circ}+\operatorname{Tg} 60^{\circ}=\ldots \) 6. \( \sin 30^{\circ} \cdot \operatorname{Cos} 60^{\circ}+\sin 45^{\circ} \cdot \operatorname{Cos} 45^{\circ}=\ldots \) 7. Buktikan \( \operatorname{Cos} 60^{\circ} \cdot \operatorname{Cos} 30^{\circ}-\sin 60^{\circ} \cdot \sin 30^{\circ}=0! \)

Berikut adalah penyelesaian dari soal-soal yang diberikan: ### 5. \( \operatorname{Tg} 30^{\circ} + \operatorname{Tg} 60^{\circ} = \ldots \) Kita tahu bahwa: \[ \operatorname{Tg} 30^{\circ} = \frac{1}{\sqrt{3}}, \quad \operatorname{Tg} 60^{\circ} = \sqrt{3} \] Maka: \[ \operatorname{Tg} 30^{\circ} + \operatorname{Tg} 60^{\circ} = \frac{1}{\sqrt{3}} + \sqrt{3} = \frac{1 + 3}{\sqrt{3}} = \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3} \] **Jawaban:** \( \frac{4\sqrt{3}}{3} \) --- ### 6. \( \sin 30^{\circ} \cdot \operatorname{Cos} 60^{\circ} + \sin 45^{\circ} \cdot \operatorname{Cos} 45^{\circ} = \ldots \) Kita hitung masing-masing komponen: \[ \sin 30^{\circ} = \frac{1}{2}, \quad \cos 60^{\circ} = \frac{1}{2} \] \[ \sin 45^{\circ} = \frac{\sqrt{2}}{2}, \quad \cos 45^{\circ} = \frac{\sqrt{2}}{2} \] Sehingga: \[ \sin 30^{\circ} \cdot \cos 60^{\circ} = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \] \[ \sin 45^{\circ} \cdot \cos 45^{\circ} = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{2}{4} = \frac{1}{2} \] \[ \frac{1}{4} + \frac{1}{2} = \frac{3}{4} \] **Jawaban:** \( \frac{3}{4} \) --- ### 7. Buktikan \( \operatorname{Cos} 60^{\circ} \cdot \operatorname{Cos} 30^{\circ} - \sin 60^{\circ} \cdot \sin 30^{\circ} = 0! \) Kita gunakan identitas trigonometri: \[ \cos(A + B) = \cos A \cos B - \sin A \sin B \] Misalkan \( A = 60^{\circ} \) dan \( B = 30^{\circ} \), maka: \[ \cos(60^{\circ} + 30^{\circ}) = \cos 90^{\circ} = 0 \] \[ \cos 60^{\circ} \cos 30^{\circ} - \sin 60^{\circ} \sin 30^{\circ} = 0 \] **Kesimpulan:** \( \operatorname{Cos} 60^{\circ} \cdot \operatorname{Cos} 30^{\circ} - \sin 60^{\circ} \cdot \sin 30^{\circ} = 0 \) terbukti benar.