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?
 A: 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, $$a^2\sin^2\gamma=c^2\sin^2\alpha$$
$$a^2(1-\cos^2\gamma)=c^2(1-\cos^2\alpha)$$
$$a^2=c^2+a^2\cos^2\gamma-c^2\cos^2\alpha$$
$$a^2=c^2+(a\cos\gamma+c\cos\alpha)(a\cos\gamma-c\cos\alpha)$$
$$a^2=c^2+b\cdot(a\cos\gamma-c\cos\alpha)$$
$$a^2=c^2+b\cdot[(b-c\cos\alpha)-c\cos\alpha]$$
$$a^2=b^2+c^2-2bc\cos\alpha$$
$\blacksquare$
A: Note that $a=2R\sin\alpha,\quad b=2R\sin\beta$ and $c=2R\sin\gamma.$ Now sbstitute these into the expression $$\dfrac{a^2+b^2-c^2}{2ab}=\dfrac{(a+b+c)(a+b-c)}{2ab}-1$$ and simplify (until you get $\cos\gamma$ :)).
A: Hint
$$a^2+b^2-c^2$$
$$=4R^2(\sin^2A+\sin^2B-\sin^2C)$$
Use Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $
and $$\sin(B+C)=\cdots=\sin A $$
A: $$\frac {b}{\sin(\gamma + \alpha)}= 2R.$$
$$\frac {b}{\sin(\gamma) \;\cos(\alpha)+ \sin\alpha \; \cos\gamma} = 2R.$$
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 $\cos^2\gamma + \sin^2\gamma =1,$
$$b^2+a^2 - 2ab\;\cos\gamma = c^2.$$
