*# PERKALIAN SINUS DAN SINUS #*
\(-2\sin\alpha \sin\beta = \cos\left ( \alpha +\beta \right )- \cos\left ( \alpha -\beta \right )\)
Atau
\(\sin\alpha \sin\beta = -\frac{1}{2} \cos\left ( \alpha +\beta \right )- \cos\left ( \alpha -\beta \right )\)
Pembuktian:
\(\cos\left ( \alpha -\beta \right )= \cos \alpha \cdot \cos\beta + \sin\alpha \cdot \sin\beta \)
\(\cos\left ( \alpha +\beta \right )= \cos \alpha \cdot \cos\beta - \sin\alpha \cdot \sin\beta \)
Kedua rumus tersebut kita eliminasi
\(\cos\left ( \alpha -\beta \right )= \cos \alpha \cdot \cos\beta + \sin\alpha \cdot \sin\beta \)
\(\underline{\cos\left ( \alpha +\beta \right )= \cos \alpha \cdot \cos\beta - \sin\alpha \cdot \sin\beta }\) __
\(\cos\left ( \alpha +\beta \right )- \cos\left ( \alpha -\beta \right )=-\left ( \sin \alpha \cdot \sin \beta \right )- \left ( \sin \alpha \cdot \sin \beta \right )\)
\(=-2\left ( \sin\alpha \cdot \sin\beta \right )\)
Maka
\(-2\sin\alpha \sin\beta = \cos\left ( \alpha +\beta \right )- \cos\left ( \alpha -\beta \right )\) Terbukti
Atau
\(2 \sin\alpha \sin\beta = \cos\left ( \alpha -\beta \right )- \cos\left ( \alpha +\beta \right )\) Terbukti
Contoh Soal
1. Hitunglah nilai dari \(\sin 45^{o} \sin 15^{o}\) adalah ...
Pembahasannya
\(\sin\alpha \sin\beta = -\frac{1}{2} \left ( \cos\left ( \alpha +\beta \right )- \cos\left ( \alpha -\beta \right ) \right )\)
\(\sin\alpha \sin\beta = -\frac{1}{2} \left ( \cos\left ( 45^{o} + 15^{o} \right )- \cos\left ( 45^{o} -15^{o} \right ) \right )\)
\(\sin\alpha \sin\beta = -\frac{1}{2} \left ( \cos 60^{o}- \cos 30^{o} \right )\)
\(\sin\alpha \sin\beta = -\frac{1}{2} \left ( \frac{1}{2} - \frac{1}{2}\sqrt{3} \right )\)
\(\sin\alpha \sin\beta = -\frac{1}{4}+\frac{1}{4}\sqrt{3}\)
2. Tentukan nilai dari \(2 \sin 52,5^{o} \sin 7,5^{o}\) adalah ...
Pembahasannya
\(2 \sin 52,5^{o} \sin 7,5^{o}= 2 \times -\frac{1}{2}\left ( \cos \left ( \alpha +\beta \right )- \cos \left ( \alpha -\beta \right ) \right )\)
\(2 \sin 52,5^{o} \sin 7,5^{o}= 2 \times -\frac{1}{2}\left ( \cos \left ( 52,5^{o}+ 7,5^{o} \right )- \cos \left ( 52,5^{o}- 7,5^{o} \right ) \right )\)
\(2 \sin 52,5^{o} \sin 7,5^{o}= - \left ( \cos \left ( 52,5^{o}+ 7,5^{o} \right )- \cos \left ( 52,5^{o}- 7,5^{o} \right ) \right )\)
\(2 \sin 52,5^{o} \sin 7,5^{o}= - \left ( \cos \left ( 60^{o} \right )- \cos \left ( 45^{o} \right ) \right )\)
\(2 \sin 52,5^{o} \sin 7,5^{o}= - \left ( \frac{1}{2}-\frac{1}{2}\sqrt{2} \right )\)
\(2 \sin 52,5^{o} \sin 7,5^{o}= \frac{1}{2}\sqrt{2}- \frac{1}{2}\)
*# PERKALIAN SINUS DAN COSINUS #*
Rumus Perkalian fungsi trigonometri sinus dan cosinus
\( 2 \cos \alpha \sin \beta = \sin \left ( \alpha +\beta \right ) - \sin \left ( \alpha -\beta \right ) \) </li>
\( 2 \sin \alpha \cos \beta = \sin \left ( \alpha +\beta \right ) + \sin \left ( \alpha -\beta \right ) \) </li>
Rumus cos dikali sin
Rumus yang akan dibahas adalah rumus cos dikali sin. Kalimat yang dapat digunakan untuk menghafal rumus perkalian cos dan sin adalah dua cos sin sama dengan sin jumlah dikurang sin selisih . Berikut rumus perkalian cos dikali sin adalah sebagai berikut:
\( 2 \cos \alpha ^{\circ} \cdot \sin \beta ^{\circ} = \sin \left ( \alpha ^{\circ} + \beta ^{\circ} \right ) - \sin \left ( \alpha ^{\circ} - \beta ^{\circ} \right ) \)
atau
\( \cos \alpha ^{\circ} \cdot \sin \beta ^{\circ} = \frac{1}{2} \left [ \sin \left ( \alpha ^{\circ} + \beta ^{\circ} \right ) - \sin \left ( \alpha ^{\circ} - \beta ^{\circ} \right ) \right ] \)
Bukti:
Pembuktian rumus cos dikali sin dapat menggunakan pengurangan rumus jumlahan dan selisih dua sudut pada fungsi sinus. perhatikan cara di bawah
\( \sin \left ( \alpha ^{\circ} + \beta \right ) = \sin \alpha ^{\circ} \cdot \cos \beta ^{\circ} + \cos \alpha ^{\circ} \cdot \sin \beta ^{\circ}\)
\( \sin \left ( \alpha ^{\circ} - \beta \right ) = \sin \alpha ^{\circ} \cdot \cos \beta ^{\circ} - \cos \alpha ^{\circ} \cdot \sin \beta ^{\circ}\) \(-\)
\(=\)
\(\sin \left ( \alpha ^{\circ} + \beta ^{\circ} \right ) - \sin \left ( \alpha ^{\circ} - \beta ^{\circ} \right ) = 2 \cos \alpha ^{\circ} \cdot \sin \beta ^{\circ} \)
atau
\( \cos \alpha ^{\circ} \cdot \sin \beta ^{\circ} = \frac{1}{2} \left [ \sin \left ( \alpha ^{\circ} + \beta ^{\circ} \right ) - \sin \left ( \alpha ^{\circ} - \beta ^{\circ} \right ) \right ] \)
Terbukti
Rumus SIN dikali COS
Rumus perkalian sinus dan cosinus yang akan dipelajari lebih lanjut adalah rumus sin dikali cos. Cara untuk menghafal perkalian sin dikali cos dapat menggunakan kalimat dua sin cos sama dengan sin jumlah ditambah sin selisih . Bentuk rumusnya adalah sebagai berikut:
\( 2 \sin \alpha ^{\circ} \cdot \cos \beta ^{\circ} = \sin \left ( \alpha ^{\circ} + \beta ^{\circ} \right )+ \sin \left ( \alpha ^{\circ} - \beta ^{\circ} \right ) \)
atau
\( \sin \alpha ^{\circ} \cdot \cos \beta ^{\circ} = \frac{1}{2} \left [ \sin \left ( \alpha ^{\circ} + \beta ^{\circ} \right ) + \sin \left ( \alpha ^{\circ} - \beta ^{\circ}\right ) \right ] \)
Bukti:
Pembuktian rumus sin dikali cos dapat menggunakan rumus penjumlah dan selisih dua sudut fungsi sinus
\( \sin \left ( \alpha ^{\circ} + \beta \right ) = \sin \alpha ^{\circ} \cdot \cos \beta ^{\circ} + \cos \alpha ^{\circ} \cdot \sin \beta ^{\circ}\)
\( \sin \left ( \alpha ^{\circ} - \beta \right ) = \sin \alpha ^{\circ} \cdot \cos \beta ^{\circ} - \cos \alpha ^{\circ} \cdot \sin \beta ^{\circ}\) \(+\)
\(=\)
\(\sin \left ( \alpha ^{\circ} + \beta ^{\circ} \right ) + \sin \left ( \alpha ^{\circ} - \beta ^{\circ} \right ) = 2 \sin \alpha ^{\circ} \cdot \cos \beta ^{\circ} \)
atau
\( \sin \alpha ^{\circ} \cdot \cos \beta ^{\circ} = \frac{1}{2} \left [ \sin \left ( \alpha ^{\circ} + \beta ^{\circ} \right ) + \sin \left ( \alpha ^{\circ} - \beta ^{\circ} \right ) \right ] \)
Terbukti
Contoh Soal
1. Diketahui besar sudut \( \alpha = 75^{\circ} \) dan sudut \( \beta = 15^{\circ}\). Maka nilai \( \cos \alpha \cdot \sin \beta \) adalah ...
Pembahasan:
Diketahui :
\(\alpha = 75^{\circ}\)
\(\beta = 15^{\circ}\)
Ditanya : \( \cos \alpha \sin \beta\)
Jawab:
\(2 \cos \alpha \sin \beta = \sin \left ( \alpha +\beta \right ) - \sin \alpha - \beta \)
\(2 \cos \alpha \sin \beta = \sin \left ( 75^{\circ} + 15^{\circ} \right ) - sin 75^{\circ} - 15^{\circ} \)
\(2 \cos \alpha \sin \beta = \cos 90^{\circ} - \cos 60^{\circ}\)
\(2 \cos \alpha \sin \beta = 1 - \frac{1}{2}\sqrt{3}\)
\(2 \cos \alpha \sin \beta = \frac{2-\sqrt{3}}{2}\)
\( \cos \alpha \sin \beta = \frac{2-\sqrt{3}}{4}\)
2. Jika \( \sin A \cos B = \frac{3}{2} \). dan \(\sin A+B = 2\). Maka \( \sin A-B = \) ...
Pembahasan :
Diketahui :
\(\sin A \cos B = \frac{3}{2} \)
\( \sin A+B = 2 \)
Ditanya : \( \sin A-B = ? \)
Jawab :
\(2 \sin A \cos B = \sin \left ( A+B \right ) + \sin \left ( A-B \right )\)
\(2 \frac{3}{2} = 2 + \sin \left ( A-B \right ) \)
\(3= 2 + \sin \left ( A-B \right ) \)
\( \sin \left ( A-B \right ) = 3-2 \)
\( \sin \left ( A-B \right ) = 1\)
