# Math Help - Prove that

1. ## Prove that

c(acosB-bcosA)=a^2-b^2

2. Originally Posted by matsci0000
c(acosB-bcosA)=a^2-b^2
Can we have the rest of the question please, is this a triangle with sides a,b,c and angles opposite the sides A, B, C?

Try the difference of the cosine rule for a and b.

CB

3. Originally Posted by matsci0000
c(acosB-bcosA)=a^2-b^2
According to cosine rule
a^2 = b^2 + c^2 - 2bc*cosA.
From this find bc*cosA. Similarly find ac*coaB, And find the difference.
Or, by using projection rule write c as
c = acosB + bcosA and simplify.