Law of cosines

From Exampleproblems

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.

1 Proof
2 Alternative proof (for acute angles)
3 Finding the angles when the sides are known
4 Isosceles case
5 See also
6 External link

Proof

Image:Vectorcosine.png

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

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

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)

Image:Law of cosines proof.png

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 b − a cos C. This makes two right triangles, one with legs a sin C and 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

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

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 $a 2 + b 2 = 2 a 2 = 2 a b$ , the law of cosines becomes