# Law of cosines

In trigonometry, the law of cosines (also known as the cosine formula) is a statement about arbitrary triangles which generalizes the Pythagorean theorem by correcting it with a term proportional to the cosine of the opposing angle. Let a, b, and c be the lengths of the sides of the triangle and let A, B, and C the respective angles opposite those sides. Then,

$c^{2}=a^{2}+b^{2}-2ab\cos C.\;$

This formula is useful for computing the third side of a triangle when two sides and their enclosed angle are known, and in computing the angles of a triangle if all three sides are known.

The law of cosines also shows that

$c^{2}=a^{2}+b^{2}\;$   if and only if   $\cos C=0.\;$

The statement cos C = 0 implies that C is a right angle, since a and b are positive. In other words, this is the Pythagorean theorem and its converse. Although the law of cosines is a broader statement of the Pythagorean theorem, it isn't a proof of the Pythagorean theorem, because the law of cosines derivation given below depends on the Pythagorean theorem.

## Proof

Using vectors and vector dot products, we can easily prove the law of cosines. If we have a triangle with vertices A, B, and C whose sides are the vectors a, b, and c, we know that:

${\mathbf {a=b-c}}\;$

and

${\mathbf {(b-c)\cdot (b-c)=b\cdot b-2b\cdot c+c\cdot c}}.\;$

Using the dot product, we simplify the above into

${\mathbf {|a|^{2}=|b|^{2}+|c|^{2}-2|b||c|}}\cos \theta .\;$

## Alternative proof (for acute angles)

Let a, b, and c be the sides of the triangle and A, B, and C the angles opposite those sides. Draw a line from angle B that makes a right angle with the opposite side b. If the length of that line is x, then sin C = x/a, which implies x = a sin C.

That is, the length of this line is a sin C. Similarly, the length of the part of b that connects the foot point of the new line and angle C is a cos C. The remaining length of b is ba cos C. This makes two right triangles, one with legs a sin C and ba cos C and hypotenuse c. Therefore, according to the Pythagorean theorem:

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

because

$\sin ^{2}C+\cos ^{2}C=1.\,$

## Finding the angles when the sides are known

By transposing the identity

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

we can find the angle C when the three sides a, b, and c are known:

$C=\arccos {\frac {a^{2}+b^{2}-c^{2}}{2ab}}.$

## Isosceles case

When $a=b$, i.e., when the triangle is isosceles with the two sides incident to the angle C equal, the law of cosines simplies significantly. Namely, because $a^{2}+b^{2}=2a^{2}=2ab$, the law of cosines becomes

$\cos C=1-{\frac {c^{2}}{2a^{2}}}.\;$