Cosine Rule Proof
We can split the triangle into two right-angled triangles like so:
(*add diagram)
Using Pythagoras we know tha:
\[①:h^2+x^2=b^2\]
and \[②:h^2+(c-x)^2=a^2\]
We can eliminate \(h\) by subtracting the equations:
\[①-②: (h^2+x^2)-(h^2+(c-x)^2)=b^2-a^2\]
\[x^2-(c-x)^2=b^2-a^2\]
\[x^2-(c^2-2cx-c^2)=b^2-a^2\]
Rearranging: \[a^2=b^2+c^2-2cx\]
We know that \(\cos A=\frac{x}{b}\) and so \(x=b \cos A\). Substituting this in gives:
\[a^2=b^2+c^2-2bc \cos A\]
Sine Rule Proof
We can split the triangle again into two right-angled triangles as we did with the cosine rule:
(* diagram)
We know that \(\sin B=\frac{h}{a}\) and \(\sin A=\frac{h}{b}\), so we see that \(h=a\sin B\) and \(h=b\sin A\).
\[\therefore a \sin B=b \sin A\]
\[\therefore \frac{a}{\sin A}=\frac{b}{\sin B}\]
Area of a Triangle Proof
Continuing to use the same triangle as before, we can easily see that the area of our triangle is \(\frac{1}{2}hc\):
(* diagram)
We can see from basic trigonometry that \(h=b\sinA\), and therefore we conclude that the area of the triangle is given by \(\frac{1}{2}bc\sinA\)