# Dari Rumus Jumlah dan Selisih Sudut

Untuk perkalian sin dengan cos rumusnya bisa didapatkan dari $\sin(\alpha + \beta)$ dan $\sin(\alpha - \beta)$

$$\begin{array}{llc} \sin(\alpha+\beta) & =\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)\\ \sin(\alpha-\beta) & =\sin(\alpha)\cos(\beta)-\cos(\alpha)\sin(\beta) & \qquad+\\ \hline \sin(\alpha+\beta)+\sin(\alpha-\beta) & =2\sin(\alpha)\cos(\beta) \end{array}$$

$$\begin{array}{llc} \sin(\alpha+\beta) & =\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)\\ \sin(\alpha-\beta) & =\sin(\alpha)\cos(\beta)-\cos(\alpha)\sin(\beta) & \qquad-\\ \hline \sin(\alpha+\beta)-\sin(\alpha-\beta) & =2\cos(\alpha)\sin(\beta) \end{array}$$

Kedua rumus di atas ekuivalen:

$$\begin{align*} 2\cos(\alpha)\sin(\beta) & =2\sin(\beta)\cos(\alpha)\\ & =\sin(\beta+\alpha)+\sin(\beta-\alpha)\\ & =\sin(\alpha+\beta)+\sin(-(\alpha-\beta))\\ & =\sin(\alpha+\beta)-\sin(\alpha-\beta) \end{align*}$$

Untuk perkalian sin dengan sin dan cos dengan cos, maka rumusnya bisa didapat dari $\cos(\alpha + \beta)$

$$\begin{array}{llc} \cos(\alpha+\beta) & =\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)\\ \cos(\alpha-\beta) & =\cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta) & \qquad+\\ \hline \cos(\alpha+\beta)+\cos(\alpha-\beta) & =2\cos(\alpha)\cos(\beta) \end{array}$$

$$\begin{array}{llc} \cos(\alpha+\beta) & =\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)\\ \cos(\alpha-\beta) & =\cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta) & \qquad-\\ \hline \cos(\alpha+\beta)-\cos(\alpha-\beta) & =-2\sin(\alpha)\sin(\beta)\\ 2\sin(\alpha)\sin(\beta) & =-(\cos(\alpha+\beta)-\cos(\alpha-\beta)) \end{array}$$

# Kesimpulan

Untuk $\alpha$ dan $\beta$ sudut apapun

$$\begin{align*} 2\sin(\alpha)\cos(\beta)&=\sin(\alpha + \beta) + \sin(\alpha -\beta)\\ 2\cos(\alpha)\sin(\beta)&=\sin(\alpha + \beta) - \sin(\alpha -\beta)\\ 2\sin(\alpha)\sin(\beta)&=-(\cos(\alpha + \beta) - \cos(\alpha -\beta))\\ 2\cos(\alpha)\cos(\beta)&=\cos(\alpha + \beta) + \cos(\alpha -\beta)\\ \end{align*}$$