en:proof-of-the-law-of-cosines

**Law of Cosines**: a^{2} = b^{2} + c^{2} - 2bccos(θ)

Divide the triangle into 2 right angle triangles:

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

Using the Pythagorean Theorem:

a^{2} = (b sin θ)^{2} + (c - b cos(θ))^{2}

= b^{2} sin ^{2} θ + c^{2} - 2cbcos(θ) + b ^{2} cos ^{2} θ

= b^{2} (sin ^{2} θ + cos ^{2} θ) + c^{2} - 2cbcos(θ)

Knowing that (sin ^{2} θ + cos ^{2} θ = 1) ^{1)}

= b^{2} * 1 + c^{2} - 2cbcos(θ)

**a ^{2} = b^{2} + c^{2} - 2cbcos(θ)**

en/proof-of-the-law-of-cosines.txt · Last modified: 2017/07/14 10:47 by federico

## Discussion