prove value for c^2 in triangle abc

Question

prove c²=a²cos(2A)+b²(cos(2B)+2ab(cos(A-B)

Answer 1

by the Law of Cosines (a more generalized Pythagorean Theorem for non right triangles)

c^2 = a^2 + b^2 -2abCosC

C=angle opposite side c, A opposite side a, and B opposite side b

C=180-A-B, as the triangle angles always sum to 180

CosC = Cos(180-A-B) = Cos(180 - (A+B))= Cos180Cos(A+B) +Sin180Sin(A-B)= -Cos(A+B) + 0=-Cos(A+B)

CosC=-Cos(A+B) substitute that into theLaw of Cosines equation

That almost gives you the last term that you're trying to prove +2abCos(A-B)

Cos2A = Cos^2(A)-Sin^2(A)=(CosA-SinA)(CosA+SinA)

Cos2B = Cos^2(B) -Sin^2(B)= (CosB-SinB)(CosB+SinB)

Cos(A+B)= CosACosB-SinASinB

1=Cos^2(A+ Sin^2(A)

1= Cos^2(B) + Sin^2(B)

which almost gives you the rest, but for a sign difference

Answer 2

Substitute Double and Difference Rule.

Combine like terms.