Law of Cosines

Law of cosines

The law of cosines (often abbreviated LoC) states:

For $\triangle ABC$ with sides $a, b, c,$ the following relationships hold: $$c^2 = a^2 + b^2 - 2ab\cos C$$ $$a^2 = b^2 + c^2 - 2ab\cos A$$ $$b^2 = c^2 + a^2 - 2ca\cos B$$ This gives us a way to find the third side of a triangle, given two sides and the angle measure between them.

Proof

We focus on proving the acute case of: $$c^2 = a^2 + b^2 - 2ab\cos C$$ As the other relationships hold the same way, just by taking different vertices. However, the obtuse triangle case must be taken separately, which is left as an exercise for the reader. :)

Notice that by right triangle trigonometry, $$AD = b\sin C$$ $$CD = b\cos C$$ Then, it follows that: $$BD = a - CD = a - b\cos C$$ By the pythagorean theorem: $$\begin{align} c^2 = AB^2 &= BD^2 + AD^2 \\ &= (a - b\cos C)^2 + (b\sin C)^2 \\ &= a^2 - 2ab\cos C + b^2\cos^2 C + b^2\sin^2 C \\ &= a^2 + b^2(\sin^2C + \cos^2C) - 2ab\cos C \end{align}$$ This almost looks like what we want, but there's an annoying $\sin^2C + \cos^2C$ in the middle. However, we note that this equals one (by a trigonometric identity), so we have: $$c^2 = a^2 + b^2 - 2ab\cos C$$ As desired.

That's it for this explanation about the law of cosines; thanks for reading!

The law of cosines simply states that $$c^2 = a^2+b^2-2ab\cos \gamma$$ in which $a,b,c,$ are side lengths, and $\gamma$ the is angle between sides $a$ and $b$ opposite to side $c$.

Now for the proof of the Law Of Cosines

From our triangle, we drop an altitude form point $C$ to side $c$, and call the foot of that altitude $D$.

so $\overline{CB}=a$,$\overline{AC}=b$, $\overline{AB}=c$, and $\angle ACB= \gamma$.Let angle $A=\alpha$ and $B= \beta$.Thus, $\overline{DB}= a \cos \beta$ and $\overline{AD}= b \cos \alpha$. Adding those together gives $$c=a \cos \beta+ b\cos \alpha.$$ Multiplying both sides by $c$ gives $$c^2=a c\cos \beta+ bc\cos \alpha \quad \text{(1)}.$$

We apply the same procedure to $\overline{AC}$ and $\overline{CB}$, writing them in terms of angles and the other sides and multiplying through by whatever side we are rewriting, to get $$\begin{align*} a^2&=a c\cos \beta+ ab\cos \gamma \quad \text{(2)}\\ b^2&=b c\cos \alpha+ ab\cos \gamma \quad \text{(3)} \end{align*}$$

Adding $\text{(2)}$ and $\text{(3)}$, we get $$a^2 + b^2 = ac\cos\beta + bc\cos\alpha + 2ab\cos\gamma \quad \text{(4)} .$$

Subtracting $\text{(1)}$ from $\text{(4)}$, we get $$a^2 + b^2 - c^2 = - ac\cos\beta - bc\cos\alpha+ ac\cos\beta + bc\cos\alpha + 2ab\cos\gamma ,$$ which simplifies down to $$c^2 = a^2 + b^2 - 2ab\cos\gamma,$$ As desired.

You are strongly encouraged to draw your own diagram so that you can play along more easily.

Here's a problem:

Consider a triangle $ABC$ with $AB=10, AC=16,$ and $\angle BAC=60^\circ.$ Find the length of $BC.$

Well, this is kinda difficult, and we don't immediately have any tools to do this.

Let

Yep, using triangle $ABD,$ we see that $BD = AB\sin BAD = 10\sin 60 = 5\sqrt{3}.$ (Why do we write this in terms of $\sin$ instead of just using 30-60-90 triangles?)

Good!

Good! We find $CD = AC - DC = 11.$ Using the Pythagorean Theorem, we get $BC^2 = BD^2 + CD^2 = (5\sqrt{3})^2 + 11^2 = 75+121 = 196$ so $BC=14.$

Alright, so we just solved our original problem. Let's try to solve this problem in general. Can you state our original problem in a more general form? For convenience, we've provided the original problem below.

Consider a triangle $ABC$ with $AB=10, AC=16,$ and $\angle BAC=60^\circ.$ Find the length of $BC.$

Here's what we have:

Consider a triangle $ABC.$ Find the length of $BC$ in terms of sides $AB,$ $AC,$ and $\angle BAC.$

Ok!

For convenience, let $|BC| = a, |AC|=b, |AB| = c,$ and $m\angle BAC = A.$ (Recall that this is fairly standard notation (link?)).

So we get, again by pythagorean theorem on $BDC$, $a^2=BC^2 = BD^2 + CD^2 \\\\ = (c\sin A)^2 + (b - c\cos A)^2 \\\\ = c^2\sin^2 A + b^2 - 2bc\cos A + c^2\cos^2 A \\\\ = b^2 - 2bc\cos A + (c^2\sin^2 A + c^2\cos^2 A) \\\\ = b^2 - 2bc\cos A + (c^2\sin^2 A + c^2\cos^2 A) \\\\ = b^2 - 2bc\cos A + c^2(\sin^2 A + \cos^2 A) \\\\ = b^2 - 2bc\cos A + c^2 \\\\$

This is the Law of Cosines. Here's some more things to think about to check your understanding.

When $A=90,$ $\cos A = 0,$ so Law of Cosines says $a^2 = b^2 + 0 + c^2$ which is just the Pythagorean theorem.

Yes, it does! When $A > 90$ $\cos A$ is negative but $D$ is also on the other side of $A$ on $AC,$ so the computation that $DC = b - c\cos A$ is still correct!

If you'd like another perspective on the Law of Cosines, see this excellent Quora answer. Enjoy!