Deriving Law of Cosines from Law of Sines 
How to eliminate $\alpha$ from the Law of Sines of plane trigonometry
$$ \dfrac{a}{\sin \alpha}= \dfrac{c}{\sin \gamma} =\dfrac{b}{\sin (\gamma+\alpha)} =2R $$
in order to arrive at the Law of Cosines
$$ c^2= a^2+b^2-2 a b \cos \gamma \;?$$
Starting to isolate $\alpha$
$$ \alpha =\sin^{-1}\big(\dfrac{b-c}{2R} +\sin  \gamma \big)-\gamma$$
$$ =\sin^{-1}( b \sin \gamma/c) -\gamma$$
involve $c$ that we are finding and so on, how best to simplify?

First, let's get the following relationship.
From $$\frac{a}{\sin\alpha}=\frac{c}{\sin\gamma}=\frac{b}{\sin(\alpha+\gamma)}$$
using componendo and dividendo,
$$\frac{a\cos\gamma+c\cos\alpha}{\sin\alpha\cos\gamma+\cos\alpha\sin\gamma}=\frac{b}{\sin(\alpha+\gamma)}$$
$$a\cos\gamma+c\cos\alpha=b$$
However, these are unnecessary stuff since obviously,

 

Now, let's go back to our problem: deriving law of cosines from law of sines.
From law of sines we have, $$a\sin\gamma=c\sin\alpha$$
Squaring, $$a^2\sin^2\gamma=c^2\sin^2\alpha$$
$$a^2(1-\cos^2\gamma)=c^2(1-\cos^2\alpha)$$
$$a^2=c^2+a^2\cos^2\gamma-c^2\cos^2\alpha$$
$$a^2=c^2+(a\cos\gamma+c\cos\alpha)(a\cos\gamma-c\cos\alpha)$$
$$a^2=c^2+b\cdot(a\cos\gamma-c\cos\alpha)$$
$$a^2=c^2+b\cdot[(b-c\cos\alpha)-c\cos\alpha]$$
$$a^2=b^2+c^2-2bc\cos\alpha$$
$\blacksquare$
