SOLUTION: prove that for any triangle ABC, where a,b, and c are respectively the sides opposite angles A,B, and C, {{{(a*cos(A) - b*cos(B)) / (b*cos(A) - a*cos(B))}}}{{{""=""}}}{{{cos(C)}}

Algebra ->  Triangles -> SOLUTION: prove that for any triangle ABC, where a,b, and c are respectively the sides opposite angles A,B, and C, {{{(a*cos(A) - b*cos(B)) / (b*cos(A) - a*cos(B))}}}{{{""=""}}}{{{cos(C)}}      Log On


   

Question 988624: prove that for any triangle ABC, where a,b, and c are respectively
the sides opposite angles A,B, and C,
%28a%2Acos%28A%29+-+b%2Acos%28B%29%29+%2F+%28b%2Acos%28A%29+-+a%2Acos%28B%29%29%22%22=%22%22cos%28C%29
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
%28a%2Acos%28A%29+-+b%2Acos%28B%29%29+%2F+%28b%2Acos%28A%29+-+a%2Acos%28B%29%29%22%22=%22%22cos%28C%29

The law of cosines: 

a%5E2+=+b%5E2%2Bc%5E2-2ab%2Acos%28A%29, b%5E2+=+a%5E2%2Bc%5E2-2ac%2Acos%28B%29, c%5E2=b%5E2%2Ba%5E2-2ab%2Acos%28C%29

when solved for the cosines, give:

cos%28A%29=%28b%5E2%2Bc%5E2-a%5E2%29%2F%282bc%29, cos%28B%29=%28a%5E2%2Bc%5E2-b%5E2%29%2F%282ac%29, cos%28C%29=%28b%5E2%2Ba%5E2-c%5E2%29%2F%282ab%29

So the identity becomes to show that 

%22%22=%22%22%28b%5E2%2Ba%5E2-c%5E2%29%2F%282ab%29

On the left side multiply numerator and denominator by LCD = 2abc





Cancel terms that add to zero and collect like terms:

+%28++a%5E2c%5E2-a%5E4-b%5E2c%5E2%2Bb%5E4+%29+%2F+%282ab%5E3-2a%5E3b%29+

Rearrange the terms of the numerator to factor by grouping, 
factor 2ab out of the denominator:

+%28+b%5E4-a%5E4-b%5E2c%5E2%2Ba%5E2c%5E2+%29+%2F+%282ab%28b%5E2-a%5E2%29%29+

Factor first two terms in numerator as the difference of squares.
Factor -c² out of last two terms in numerator:

 

Factor (b²-a²) out of numerator:



+%28+%28b%5E2-a%5E2%29%28b%5E2%2Ba%5E2-c%5E2%29+%29+%2F+%282ab%28b%5E2-a%5E2%29%29+

Cancel (b²-a²)'s

+%28b%5E2%2Ba%5E2-c%5E2%29+%2F+%282ab%29+

That is the right side of the identity.

Edwin