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# Law of cosines

In trigonometry, the law of cosines 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 sides of the triangle and A, B, and C the 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's values are known, and in computing the angles of a triangle if all three sides' values are known.

The law of cosines also shows that

$c^2 = a^2 + b^2 \;$ iff $\cos C = 0 . \;$

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.

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## Derivation (for acute angles)

Triangle

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 $\sin C = \frac{x}{a} , \;$ 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 $b - a \cos C. \;$ This makes two right triangles, one with legs $a \sin C , \;$ $b - a \cos C , \;$ and hypotenuse c. Therefore, according to the Pythagorean theorem:

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

because

$\sin^2 C + \cos^2 C=1. \;$

## Law of cosines using vectors

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:

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

since

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

Using the dot product, we simplify this into

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