====== Proof of The Law of Cosines ======

{{:en:law-of-cosines1-fs8.png?nolink|}}

**Law of Cosines**: a<sup>2</sup> = b<sup>2</sup> + c<sup>2</sup> - 2bccos(θ)

===== Proof =====

Divide the triangle into 2 right angle triangles:

{{:en:law-of-cosines2-fs8.png?nolink|}}

Using trigonometry, the sides of the green triangle are 
  * Longest side: **a**
  * Side on the left: **b . sin θ** ; because the sine of θ is opposite divided the hypotenuse, and the opposite is the length of this side. Then, the length of this side is the hypotenuse multiplied by the sine of θ  
  * Side on the right: ***c - b cos(θ)** ; we must subtract from the length "c", the part of c that is occupied by the red triangle. We know that, in the red triangle, cos θ is the adjacent side divided the hypotenuse. The adjacent is the "base" of the red triangle, and it is cos θ multiplied by the hypotenuse. The hypotenuse is b. Then, the base of the red triangle is (cos θ)b. So, the length of the "base" of the green triangle is c - b.cos θ

{{:en:law-of-cosines3-fs8.png?nolink|}}

Using the Pythagorean Theorem:

a<sup>2</sup> = (b sin θ)<sup>2</sup> + (c - b cos(θ))<sup>2</sup>

= b<sup>2</sup> sin <sup>2</sup> θ + c<sup>2</sup> - 2cbcos(θ) + b <sup>2</sup> cos <sup>2</sup> θ

= b<sup>2</sup> (sin <sup>2</sup> θ + cos <sup>2</sup> θ) + c<sup>2</sup> - 2cbcos(θ)

Knowing that (sin <sup>2</sup> θ + cos <sup>2</sup> θ = 1) ((see https://en.wikibooks.org/wiki/Trigonometry/Sine_Squared_plus_Cosine_Squared))

= b<sup>2</sup> * 1 + c<sup>2</sup> - 2cbcos(θ)

**a<sup>2</sup> = b<sup>2</sup> + c<sup>2</sup> - 2cbcos(θ)**