Trigonometry/Law of Cosines

Law of Cosines

The Pythagorean theorem is a special case of the more general theorem relating the lengths of sides in any triangle, the law of cosines:

$a^2+b^2-2ab\cos{\theta}=c^2, \,$

where θ is the angle between sides a and b.

Does the formula make sense?

This formula had better agree with the Pythagorean Theorem when $\displaystyle \theta=90^\circ$.

So try it...

When $\displaystyle \theta=90^\circ$, $\displaystyle \cos\theta=\cos 90^\circ=0$

The $\displaystyle -2ab\cos\theta=0$ and the formula reduces to the usual Pythagorean theorem.

Permutations

For any triangle with angles A, B, and C and corresponding opposite side lengths a, b, and c, the Law of Cosines states that

$a^2=b^2 + c^2 - 2bc \cdot\cos A,$

$b^2=a^2 + c^2 - 2ac \cdot\cos B,$

$c^2=a^2 + b^2 - 2ab \cdot\cos C.$

Proof

Dropping a perpendicular OC from vertex C to intersect AB (or AB extended) at O splits this triangle into two right-angled triangles AOC and BOC, with altitude h from side c.

First we will find the lengths of the other two sides of triangle AOC in terms of known quantities, using triangle BOC.

h=a sin B

Side c is split into two segments, with total length c.

OB has length a cos B

AO has length c - a cos B

Now we can use the Pythagorean Theorem to find b, since b² = AO² + h².

$\begin{matrix}b^2 & = & (c-a\cos B)^2+a^2\sin^2 B \\ \ &=& c^2-2ac\cos B +a^2\cos^2 B+a^2\sin^2 B\\ \ &=& a^2+c^2-2ac\cos B\end{matrix}$

The corresponding expressions for a and c can be proved similarly.

The formula can be rearranged:

$cos(C) = \frac{a^2+b^2-c^2}{2ab}$

and similarly for cos(A) and cos(B).

Applications

This formula can be used to find the third side of a triangle if the other two sides and the angle between them are known. The rearranged formula can be used to find the angles of a triangle if all three sides are known. See Solving Triangles Given SAS.