Triangle $ABC$ of sides $a$ and $b$ - Prove $\frac{a\cos A + b\cos B}{a\cos B + b\cos A} = \cos(A - B)$

Question

I cannot solve this:

Using the usual notation in $\triangle ABC$, prove that $$\frac{a\cos A + b\cos B}{a\cos B + b\cos A} = \cos(A - B)$$

I know that $b\cos A + a\cos B = c$ but cannot find any meaningful connection between the RHS & LHS

I know the sine and cosine rules but cannot see how they might apply

Answer 1

Apply the sine formula $\frac ab = \frac{\sin A}{\sin B}$,

$$\frac{a\cos A + b\cos B}{a\cos B + b\cos A} = \frac{\sin A\cos A + \sin B\cos B}{\sin A\cos B + \sin B\cos A} $$ $$= \frac{\sin 2A + \sin 2B}{2\sin (A+ B )}= \frac{2\sin(A+B)\cos(A-B)}{2\sin (A+ B )}=\cos(A - B)$$

Answer 2

Hint: Use that $$\cos(A)=\frac{b^2+c^2-a^2}{2bc}$$ etc and $$\cos(A-B)=\cos(A)\cos(B)+\sin(A)\sin(B)$$ and $$S=\frac{1}{2}bc\sin(A)=\frac{1}{2}ac\sin(B)$$ and $$S=\sqrt{p(p-a)(p-b)(p-c)}$$ where $$p=\frac{a+b+c}{2}$$

Answer 3

Use https://en.m.wikipedia.org/wiki/Law_of_sines to find $$\dfrac{\sin2A+\sin2B}{2(\sin A\cos B+\cos A\sin B)}$$

Use http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html in the numerator

and can you recognize the formula for the denominator