Proof of the law of cosines

One of the most elegant proofs of the law of cosines uses vectors and the scalar product.

Proof — Via the scalar product

We interpret the sides as vectors: $\vec{AB} = \vec{c}$, $\vec{AC} = \vec{b}$, $\vec{BC} = \vec{a}$. Chasles’ relation holds: $\vec{a} = \vec{BC} = \vec{AC} - \vec{AB} = \vec{b} - \vec{c}.$ We compute the squared modulus using the scalar product: $|\vec{a}|^2 = \vec{a}\cdot\vec{a} = (\vec{b}-\vec{c})\cdot(\vec{b}-\vec{c}) = |\vec{b}|^2 - 2\,\vec{b}\cdot\vec{c} + |\vec{c}|^2.$ But $\vec{b}\cdot\vec{c} = |\vec{b}|\,|\vec{c}|\cos\widehat{A}$, because the angle between $\vec{AB}$ and $\vec{AC}$ is precisely $\widehat{A}$. Substituting and using $|\vec{a}|=a$, $|\vec{b}|=b$, $|\vec{c}|=c$: $a^2 = b^2 + c^2 - 2bc\cos\widehat{A}. \qquad \blacksquare$

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Topics: Triangle trigonometry
Concepts: Scalar product · The law of cosines
Functions: Cosine
Skills: Proving