I was reading a book. Where I found some equations.
$$\sin (\alpha+\beta)=\sin \alpha \cos \beta + \cos \alpha \sin \beta$$
$$\sin (\alpha+\beta)=\cos \alpha \cos \beta - \sin \alpha \sin \beta$$
$$\sin (\alpha-\beta)=\sin \alpha \cos \beta - \cos \alpha \sin \beta$$
$$\sin (\alpha-\beta)=\cos \alpha \cos \beta + \sin \alpha \sin \beta$$
I could derive double angle formula from them.
$$\sin (2\alpha)=\sin \alpha \cos \alpha+\cos \alpha\sin \alpha$$
$$=2\sin\cos\alpha$$
Same way I had derived for $\cos$ also. But, I didn't find how to prove that.
While, $$\sin (\alpha+\beta)\ne \sin \alpha + \sin \beta$$
than, I can't use algebra to derive them. Even, some old trigonometric formulas was derived using a simple triangle. But, here they have given two angle that's why I can't understand how to solve it.
