Question

In the triangle OAB, OA = a and OB= b. Show that (a-b) times (a-b) = (a times a) + (b times b) - (2 times a times b). Hence prove the cosine rule

Specifically the Cosine Rule part of question is difficult

Thanks

Answer 1

Let's say that θ = angle AOB, and that θ is acute.
Drop an altitude from B to OA. Let h be its length
Using the Pythagorean Theorem,
(b cosθ)^2 + h^2 = b^2

Let c be the length of side AB
(a - bcosθ)^2 + h^2 = c^2

Eliminating h, we have

b^2 - (b cosθ)^2 = c^2 - (a - bcosθ)^2
b^2 - b^2 cos^2θ = c^2 - a^2 + 2ab cosθ - b^2 cos^2θ
c^2 = a^2 + b^2 - 2ab cosθ

You can do similar gymnastics if θ is obtuse.