If the sides of $\triangle ABC$ are $a = 4$, $b = 6$ and $c = 8$, then show that $4 \cos B + 3 \cos C = 2$
By cosine rule, $\;$ $\cos B = \dfrac{c^2 + a^2 - b^2}{2 c a}$, $\;$ $\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$
Given: $\;$ $a = 4$, $b = 6$, $c = 8$
$\therefore \;$ $\cos B = \dfrac{8^2 + 4^2 - 6^2}{2 \times 8 \times 4} = \dfrac{64 + 16 - 36}{64} = \dfrac{11}{16}$
$\cos C = \dfrac{4^2 + 6^2 - 8^2}{2 \times 4 \times 6} = \dfrac{16 + 36 - 64}{48} = - \dfrac{1}{4}$
$\therefore \;$ $LHS = 4 \cos B + 3 \cos C = 4 \times \dfrac{11}{16} + 3 \times \left(- \dfrac{1}{4}\right) = \dfrac{11}{4} - \dfrac{3}{4} = 2 = RHS$
Hence proved.