Identitas Perkalian Trigonometri

 $\sin\ \alpha\ \cos\ \beta\ =\ \frac{1}{2}\left\{\sin\ \left(\alpha+\beta\right)+\sin\ \left(\alpha-\beta\right)\right\}$  

 $\cos\ \alpha\ \sin\ \beta\ =\ \frac{1}{2}\left\{\sin\ \left(\alpha+\beta\right)-\sin\ \left(\alpha-\beta\right)\right\}$  

 $\cos\ \alpha\ \cos\ \beta\ =\frac{1}{2}\left\{\cos\ \left(\alpha+\beta\right)+\cos\ \left(\alpha-\beta\right)\right\}$  

 $\sin\ \alpha\ \sin\ \beta\ =-\frac{1}{2}\left\{\cos\ \left(\alpha+\beta\right)-\cos\ \left(\alpha-\beta\right)\right\}$  

 $\sin\ 45\degree\ \sin\ 15\degree=-\frac{1}{2}\left\{\cos\ \left(45+15\right)\degree-\cos\ \left(45-15\right)\degree\right\}$  

 $\sin\ 45\degree\ \sin\ 15\degree=-\frac{1}{2}\left\{\cos\ \left(60\right)\degree-\cos\ \left(30\right)\degree\right\}$  

 $\sin\ 45\degree\ \sin\ 15\degree=-\frac{1}{2}\left\{\frac{1}{2}-\frac{1}{2}\sqrt{3}\right\}$  

 $\sin\ 45\degree\ \sin\ 15\degree=-\frac{1}{4}+\frac{1}{4}\sqrt{3}$  

 $\sin\ 45\degree\ \sin\ 15\degree=\frac{1}{4}\sqrt{3}-\frac{1}{4}$ 

 $\cos\ 75\degree\ \cos\ -15\degree=\frac{1}{2}\left\{\cos\ \left(75+\left(-15\right)\right)\degree+\cos\ \left(75-\left(-15\right)\right)\degree\right\}$  

 $\cos\ 75\degree\ \cos\ -15\degree=\frac{1}{2}\left\{\cos\ \left(60\right)\degree+\cos\ \left(90\right)\degree\right\}$  

 $\cos\ 75\degree\ \cos\ -15\degree=\frac{1}{2}\left\{\frac{1}{2}+0\right\}$  

 $\cos\ 75\degree\ \cos\ -15\degree=\frac{1}{4}$