So I'm trying to understand a law of cosines proof that involves the distance formula and I'm having trouble. I've included the proof below from wikipedia that I'm trying to follow. What I'm have trouble understanding is the way they define the triangle point A.
I've always been taught that cosine represents $\frac{adj}{hyp}$ so I'm not sure what cosine represents outside the context of a right triangle. After I understand this I can follow the proof I'm just trying to understand how $b \cos\gamma,\ b \sin\gamma$ represent the x and y coordinates with a generic angle in $\gamma$. Any other resources or advice would be appreciated.
$A = (b \cos\gamma,\ b \sin\gamma),\ B = (a,\ 0),\ \text{and}\ C = (0,\ 0)\,.$
By the distance formula, we have
$c = \sqrt{(a – b \cos\gamma)^2 + (0 – b \sin\gamma)^2}\,.$
Now, we just work with that equation:
:\begin{align}
c^2 & {} = (a – b \cos\gamma)^2 + (- b \sin\gamma)^2 \\
c^2 & {} = a^2 – 2 a b \cos\gamma + b^2 \cos^2 \gamma + b^2 \sin^2 \gamma \\
c^2 & {} = a^2 + b^2 (\sin^2 \gamma + \cos^2 \gamma) – 2 a b \cos\gamma \\
c^2 & {} = a^2 + b^2 – 2 a b \cos\gamma\,.
\end{align}
Best Answer
$c$ is the length of the line $AB$, i.e. from $(b \cos\gamma,\ b \sin\gamma)$ to $(a,\ 0)$, which has horizontal component $a - b \cos\gamma$ and vertical component $0 - b \sin\gamma$.
So using Pythagoras, $c = \sqrt{(a - b \cos\gamma)^2 + (0 - b \sin\gamma)^2}$.
Then expand, and simplify using Pythagoras again in $\sin^2 \gamma + \cos^2 \gamma =1$.