Derive$~\frac{1}{2}\left\{\sin\left(2\alpha+\beta\right)+\sin\left(\beta\right)\right\}~$from$~\sin\left(\alpha+\beta\right)\cos\left(\alpha\right)$

Question

I want to derive the below RHS from the below LHS .

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

Especially, the coefficient of $~ 2 ~$ of $~ \sin^{}\left(2 \alpha + \beta\right) ~$ is currently the bigget part of the problems .

I've only came up of $$~ \sin^{}\left(\alpha + \beta\right) = \sin^{}\left(\alpha \right) \cos^{}\left(\beta\right) + \cos^{}\left(\alpha \right) \sin^{}\left(\beta\right) ~$$

Which formula(s) can be used at here to derive the RHS from the LHS?

ADD

$$ \sin^{}\left(\alpha \right) = \frac{ \exp\left(j \alpha \right) - \exp\left(-j \alpha \right) }{ 2j } $$

$$ \cos^{}\left(\alpha \right) = \frac{ \exp\left(j \alpha \right) + \exp\left(- j \alpha \right) }{ 2 } $$

$$ \sin^{}\left(2 \alpha + \beta\right) $$

$$ = \sin^{}\left(\alpha + \left( \alpha + \beta \right) \right)= \sin^{}\left(\alpha \right) \cos^{}\left(\alpha + \beta\right) + \cos^{}\left(\alpha \right) \sin^{}\left(\alpha + \beta\right) $$

$$ \therefore ~~ \sin^{}\left(2 \alpha + \beta\right) - \sin^{}\left(\alpha \right) \cos^{}\left(\alpha + \beta\right) = \underbrace{\cos^{}\left(\alpha \right) \sin^{}\left(\alpha + \beta\right)}_\text{LHS which I firstly wrote .} $$

Answer 1

$$\sin(\alpha+\beta)\cos(\alpha)=\sin(\alpha)\cos(\beta)\cos(\alpha)+\cos(\alpha)\sin(\beta)\cos(\alpha)=\sin(\alpha)\cos(\alpha)\cos(\beta)+\cos^2(\alpha)\sin(\beta)=\frac12\sin(2\alpha)\cos(\beta)+\cos^2(\alpha)\sin(\beta)=\frac12\sin(2\alpha)\cos(\beta)+\frac12\cos(2\alpha)\sin(\beta)+\frac12\sin(\beta)=\frac12\sin(2\alpha+\beta)+\frac12\sin(\beta)=\frac12(\sin(2\alpha+\beta)+\sin(\beta))$$

Progressing from between steps requires some algebra and the following identities: $$\sin(A+B)=\sin(A)\cos(B)+\cos(A)\sin(B)$$ $$\sin(2A)=2\sin(A)\cos(A)$$ $$\cos^2(A)=\frac12+\frac12\cos(2A)$$