Properties of Triangle - Trigo Problem : In $\triangle $ABC prove that $a\cos(C+\theta) +c\cos(A-\theta) = b\cos\theta$

Question

Problem :

In $\triangle $ABC prove that $a\cos(C+\theta) +\cos(A-\theta) = b\cos\theta$

My approach :

Using $\cos(A+B) =\cos A\cos B -\sin A\sin B and \cos(A-B) = \cos A\cos B +\sin A\sin B$, we get: \begin{equation} a(\cos C\cos\theta -\sin C\sin\theta) +c(\cos A\cos\theta +\sin A\sin\theta)\quad\\=\cos\theta(a\cos C+c\cos A) +\sin\theta(c\sin A -a\sin C)\qquad(\text{i}) \end{equation}

By using projection formula which states: $b=a\cos C +c \cos A $

(i) will become :

$$b\cos\theta +\sin\theta (c\sin A -a\sin C)$$ How do I proceed from here?

Answer 1

Law of sines states that

$$\frac{\sin{A}}{a} = \frac{\sin{C}}{c}$$

which means that $c \sin{A} - a \sin{C} = 0$. The desired result follows.