In $\triangle ABC$, if $\theta$ be any angle, then prove that $\;$ $b \cos \theta = c \cos \left(A - \theta\right) + a \cos \left(C + \theta\right)$
In $\triangle ABC$, we have by sine rule,
$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$
$\implies$ $\sin A = \dfrac{a}{2 R}$, $\;$ $\sin B = \dfrac{b}{2 R}$, $\;$ $\sin C = \dfrac{c}{2 R}$
By projection formula we have,
$b = c \cos A + a \cos C$
Now,
$\begin{aligned}
RHS & = c \cos \left(A - \theta\right) + a \cos \left(C + \theta\right) \\\\
& \left[\text{Note: } \cos \left(\alpha - \beta\right) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \right. \\
& \left. \hspace{1.1cm} \cos \left(\alpha + \beta\right) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \right] \\\\
& = c \left(\cos A \cos \theta + \sin A \sin \theta\right) + a \left(\cos C \cos \theta - \sin C \sin \theta\right) \\\\
& = \cos \theta \left(a \cos C + c \cos A\right) + \sin \theta \left(c \sin A - a \sin C\right) \\\\
& = b \cos \theta + \sin \theta \left(c \times \dfrac{a}{2 R} - a \times \dfrac{c}{2 R}\right) \;\; \left[\text{by projection formula and sine rule}\right] \\\\
& = b \cos \theta = LHS
\end{aligned}$
Hence proved.