Q:

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:

$$\frac {b}{sin(\gamma + \alpha)}= 2R$$
$$\frac {b}{sin\gamma \;cos\alpha+ sin\alpha \; cos\gamma} = 2R$$
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
$$b= c\;\sqrt{1-\frac{a^2\;sin^2\gamma}{c^2}} + a\;cos\gamma \;=\; \sqrt{c^2-a^2\;sin^2\gamma} + a\;cos\gamma$$
$$(b-a\;cos\gamma)^2 = c^2-a^2\;sin^2\gamma$$
$$b^2+a^2cos^2\gamma - 2ab\;cos\gamma = c^2-a^2\;sin^2\gamma$$
$$b^2+a^2cos^2\gamma - 2ab\;cos\gamma + a^2\;sin^2\gamma = c^2$$
$$b^2+a^2(cos^2\gamma + sin^2\gamma) - 2ab\;cos\gamma = c^2$$
Since $cos^2\gamma + sin^2\gamma =1$
Then
$$b^2+a^2 - 2ab\;cos\gamma = c^2$$

