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?

$$\frac {b}{\sin(\gamma + \alpha)}= 2R.$$
$$\frac {b}{\sin(\gamma) \;\cos(\alpha)+ \sin\alpha \; \cos\gamma} = 2R.$$
From Law of Sines: $\sin\alpha = \frac {a}{2R} \;\; and\;\; \sin(\gamma)= \frac {c}{2R}.$
Then
$$\frac {b}{\frac{c}{2R} \;\cos\alpha+ \frac{a}{2R} \; \cos\gamma} = 2R,$$
$$b= c\;\cos\alpha + a\;\cos\gamma,$$
$$b= c\;\sqrt{1-\sin^2\alpha} + a\;\cos\gamma.$$
From Law of Sines: $\sin\alpha = \frac{a\;\sin\gamma}{c}.$
Then
$$b= c\;\sqrt{1-\frac{a^2\;\sin^2\gamma}{c^2}} + a\;\cos\gamma \;=\; \sqrt{c^2-a^2\;\sin^2\gamma} + a\;\cos\gamma$$
$$(b-a\;\cos\gamma)^2 = c^2-a^2\;\sin^2\gamma,$$
$$b^2+a^2\cos^2\gamma - 2ab\;\cos\gamma = c^2-a^2\;\sin^2\gamma,$$
$$b^2+a^2\cos^2\gamma - 2ab\;\cos\gamma + a^2\;\sin^2\gamma = c^2,$$
$$b^2+a^2(\cos^2\gamma + \sin^2\gamma) - 2ab\;\cos\gamma = c^2.$$
Since $\cos^2\gamma + \sin^2\gamma =1,$
$$b^2+a^2 - 2ab\;\cos\gamma = c^2.$$
