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Subsection B.4.1 Cosine Law or Law of Cosines

The cosine law says that, if a triangle has sides of length \(a\text{,}\) \(b\) and \(c\) and the angle opposite the side of length \(c\) is \(\gamma\text{,}\) then

\begin{align*} c^2 &= a^2+b^2 - 2ab\cos\gamma \end{align*}

Observe that, when \(\gamma=\tfrac{\pi}{2}\text{,}\) this reduces to, (surpise!) Pythagoras' theorem \(c^2=a^2+b^2\text{.}\) Let's derive the cosine law.

Consider the triangle on the left. Now draw a perpendicular line from the side of length \(c\) to the opposite corner as shown. This demonstrates that

\begin{align*} c &= a \cos \beta + b \cos \alpha\\ \end{align*}

Multiply this by \(c\) to get an expression for \(c^2\text{:}\)

\begin{align*} c^2 &= ac \cos \beta + bc \cos \alpha\\ \end{align*}

Doing similarly for the other corners gives

\begin{align*} a^2 &= ac \cos \beta + ab \cos \gamma\\ b^2 &= bc \cos \alpha + ab \cos \gamma \end{align*}

Now combining these:

\begin{align*} a^2+b^2-c^2 &= (bc-bc) \cos \alpha + (ac-ac)\cos\beta + 2ab \cos \gamma\\ &= 2ab\cos \gamma \end{align*}

as required.