[Boutique] Proof Cosine Theorem

Part 1: Proving the cosine theorem

Ma=√^2-ac*cosB)

=√

By b^2=a^2+c^2-2ac*cosB

Let, 4ac*cosB=2a^2+2c^2-2b^2, substitute the above ma expression:

Ma=√[4c^2+a^2-]

=√

The certificate is completed.

Part 2: Proving the cosine theorem

Cosine theorem proving process ma=√^2-ac*cosB)

=√

By b^2=a^2+c^2-2ac*cosB

Let, 4ac*cosB=2a^2+2c^2-2b^2, substitute the above ma expression:

Ma=√[4c^2+a^2-]

=√

The certificate is completed.

Part 3: Proving the cosine theorem

In any ΔABC, AD⊥BC.

∠C is the opposite side of c, ∠B is the opposite side of b, and ∠A is the opposite side of a -->

BD=cosB*c, AD=sinB*c, DC=BC-BD=a-cosB*c

The Pythagorean Theorem knows:

AC2=AD2+DC2

B2=2+2

B2=sin2B*c2+a2+cos2B*c2-2ac*cosB

B2=*c2-2ac*cosB+a2

B2=c2+a2-2ac*cosB

So, cosB=/2ac