Timeless Theorems of Mathematics/Law of Cosines

Here $c^2=a^2+b^2-2ab\cdot \cos \gamma$

The law of cosines explains the relation between the sides and an angle of a triangle. The law states that for any triangle $\mathrm{\Delta }ABC$, if $\mathrm{\angle }ACB=\theta$, $AB=c$, $AC=b$ and $BC=a$, then $c^2=a^2+b^2-2ab\cdot \cos \theta$. For $\theta =\frac{\pi }{2},\mathrm{\Delta }ABC$ is a right angle and $c^2=a^2+b^2$, we have proved it as the pythagorean theorem earlier.

Proof

Statement

In any triangle, the square of one side's length is equal to the difference between the sum of the squares of the other two sides' lengths and twice the product of those two sides' lengths and the cosine of the included angle.

Assume, $\mathrm{\Delta }ABC$ is a triangle where $\mathrm{\angle }ACB=\theta ,$ $AB=c$ $AC=b$ and $BC=a.$ It is needed to be proved that, $c^2=a^2+b^2-2ab\cdot \cos \theta$.

Let us extend the line segment $BC$ to $BD$ (only for obtuse triangles), where $AD\perp BD$. Assume $AD=h$ and $CD=k$.

Proof with the help of the Pythagorean Theorem

For obtuse triangles:

Construction for proving the law of cosines for angle $\theta >\pi /2$.

According to the Pythagorean theorem, we can say that, for $\mathrm{\Delta }ACD,$ $b^2=k^2+h^2$

Similarly, for the triangle $\mathrm{\Delta }ABD,$ $c^2=h^2+(a+k{)}^2$ $=h^2+a^2+k^2+2ak$ $=a^2+b^2+2ak$.

We will be using this value for further proof. But now, let's determine some trigonometric values for the triangles.

Here, $\mathrm{\angle }ACD=\pi -\theta$.

Therefore, $\frac{k}{b}=\cos (\pi -\theta )$

Or, $k=b\cdot \cos (\pi -\theta )$ $=b[\cos (\pi )\cdot cos(\theta )+\sin (\pi )\cdot \sin (\theta )]$ $=-b\cdot \cos (\theta )$

Now, $c^2=a^2+b^2+2ak$.

$\therefore c^2=a^2+b^2-2ab\cdot \cos \theta$. [Proved]

For acute angles:

Construction for proving the law of cosines for angle $\theta <\pi /2$.

Like as the proof we have proved before, according to the Pythagorean theorem, we can say that, for $\mathrm{\Delta }ACD,$ $b^2=k^2+h^2$

Similarly, for the triangle $\mathrm{\Delta }ABD,$ $c^2=h^2+(a-k{)}^2$ $=h^2+a^2+k^2-2ak$ $=a^2+b^2-2ak$.

Here, $\mathrm{\angle }ACD=\theta$.

Therefore, $\frac{k}{b}=\cos \theta$ $\Rightarrow k=b\cdot \cos \theta$

Now, $c^2=a^2+b^2-2ak$.

$\therefore c^2=a^2+b^2-2ab\cdot \cos \theta$. [Proved]

[Note: Whatever the triangle is, the formula, $c^2=a^2+b^2-2ab\cdot \cos \theta$ works.]