Proof of Law of Cosines

Proof of the Law of Cosines

The Law of Cosines is given by the equation:

$$ c^2 = a^2 + b^2 - 2ab \cos(\theta) $$

Consider a triangle with sides \(a\), \(b\), and \(c\), and an angle \(\theta\) opposite side \(c\).

Derivation Steps:

1. Triangle Setup:

   Position the triangle in the Cartesian plane with vertices at \((0, 0)\), \((a, 0)\), and \((x, y)\).

2. Vertex Coordinates:

   Using trigonometry, the coordinates of the third vertex are:

   • \( x = b \cos(\theta) \)
   • \( y = b \sin(\theta) \)
3. Applying the Distance Formula:

   The length of side \(c\) is found by:

   $$ c = \sqrt{(x - a)^2 + y^2} $$

4. Coordinate Substitution:

   Insert \( x = b \cos(\theta) \) and \( y = b \sin(\theta) \) into the distance formula:

   $$ c = \sqrt{(b \cos(\theta) - a)^2 + (b \sin(\theta))^2} $$

5. Expansion:

   Expand the square root expression:

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

6. Pythagorean Identity:

   Utilize \( \cos^2(\theta) + \sin^2(\theta) = 1 \):

   $$ c = \sqrt{a^2 + b^2 - 2ab \cos(\theta)} $$

7. Squaring:

   Eliminate the square root by squaring both sides:

   $$ c^2 = a^2 + b^2 - 2ab \cos(\theta) $$