Proof of the Second Cosine Law Using the Definition of Trigonometric Functions
Formula
For the triangle given above, the following equations hold true, and they are collectively known as the law of cosines.
$$ \begin{cases} a^{2} =b^{2}+c^{2}-2bc\cos\alpha \\ b^{2}=a^{2}+c^{2}-2ac\cos\beta \\ c^{2}=a^{2}+b^{2}-2ab\cos\gamma \end{cases} $$
Proof
From the triangle in the upper left corner of the diagram, we obtain the following equation.
$$ \begin{align} a &= \overline{BH_{a}}+\overline{H_{a}C} \nonumber \\ &= c\cos\beta + b\cos\gamma \label{eq1} \end{align} $$
Multiplying both sides by $a$ yields:
$$ a^{2}=ac\cos\beta + ab\cos \gamma $$
The same applies to $b$ and $c$, resulting in:
$$ \begin{align} && b &= \overline{AH_{b}}+\overline{H_{b}C} \nonumber \\ && &= c\cos\alpha + a\cos\gamma \label{eq2} \\ && \implies b^{2}&=bc\cos\alpha + ab\cos\gamma \nonumber \end{align} $$
$$ \begin{align} &&c &= \overline{AH_{c}}+\overline{H_{c}B} \nonumber \\ && &= b\cos\alpha + a\cos\beta \label{eq3} \\ \implies &&c^{2}&=bc\cos\alpha + ac\cos\beta \nonumber \end{align} $$
Therefore, we obtain the following result:
$$ \begin{align*} b^{2}+c^{2} &= (bc\cos\alpha + ab\cos\gamma) + (bc\cos\alpha + ac\cos\beta) \\ &= (ab\cos\gamma+ ac\cos\beta)+2bc\cos\alpha \\ &= a^{2}+2bc\cos\alpha \end{align*} $$
Rearranging $a^{2}$ gives:
$$ a^{2}=b^{2}+c^{2}-2bc\cos\alpha $$
The same method can be used to prove $b^{2}$ and $c^{2}$.
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$\eqref{eq1}$, $\eqref{eq2}$, $\eqref{eq3}$ are collectively referred to as the first law of cosines.