# Dari Rumus Perkalian Sin dan Cos

Rumus ini bisa diturunkan dari rumus perkalian sinus dan cosinus

Misal $\alpha + \beta = A$ dan $\alpha - \beta = B$

$$\begin{array}{llc} \alpha+\beta & =A\\ \alpha-\beta & =B & +\\ \hline 2\alpha & =A+B\\ \alpha & =\frac{1}{2}(A+B) \end{array}$$

$$\begin{array}{llc} \alpha+\beta & =A\\ \alpha-\beta & =B & -\\ \hline 2\beta & =A-B\\ \beta & =\frac{1}{2}(A-B) \end{array}$$

$$\begin{align*} 2\sin(\alpha)\cos(\beta) & =\sin(\alpha+\beta)+\sin(\alpha-\beta)\\ 2\sin(\frac{1}{2}(A+B))\cos(\frac{1}{2}(A-B)) & =\sin(A)+\sin(B) \end{align*}$$

$$\begin{align*} 2\cos(\alpha)\sin(\beta) & =\sin(\alpha+\beta)-\sin(\alpha-\beta)\\ 2\cos(\frac{1}{2}(A+B))\sin(\frac{1}{2}(A-B)) & =\sin(A)-\sin(B) \end{align*}$$

$$\begin{align*} 2\sin(\alpha)\sin(\beta) & =-(\cos(\alpha+\beta)-\cos(\alpha-\beta))\\ 2\sin(\frac{1}{2}(A+B))\sin(\frac{1}{2}(A-B)) & =-(\cos(A)-\cos(B))\\ -2\sin(\frac{1}{2}(A+B))\sin(\frac{1}{2}(A-B)) & =\cos(A)-\cos(B) \end{align*}$$

$$\begin{align*} 2\cos(\alpha)\cos(\beta) & =\cos(\alpha+\beta)+\cos(\alpha-\beta)\\ 2\cos(\frac{1}{2}(A+B))\cos(\frac{1}{2}(A-B)) & =\cos(A)+\cos(B) \end{align*}$$

# Kesimpulan

Untuk A dan B sudut apapun

$$\begin{align*} \sin(A)+\sin(B)&=2\sin(\frac{1}{2}(A+B))\cos(\frac{1}{2}(A-B))\\ \sin(A)-\sin(B)&=2\cos(\frac{1}{2}(A+B))\sin(\frac{1}{2}(A-B))\\ \cos(A)-\cos(B)&=-2\sin(\frac{1}{2}(A+B))\sin(\frac{1}{2}(A-B))\\ \cos(A)+\cos(B)&=2\cos(\frac{1}{2}(A+B))\cos(\frac{1}{2}(A-B))\\ \end{align*}$$