Help with trigonometric proof

Question

Show that $\frac {\cos (24)}{\cos {6} }+2\times \sin {24}=\sqrt {3}$

Answer 1

We need to prove that $$\cos24^{\circ}+2\sin24^{\circ}\cos6^{\circ}=2\sin60^{\circ}\cos6^{\circ}$$ or $$\cos24^{\circ}+\sin30^{\circ}+\sin18^{\circ}=\sin66^{\circ}+\sin54^{\circ}$$ or $$\sin66^{\circ}+\frac{1}{2}+\sin18^{\circ}=\sin66^{\circ}+\sin54^{\circ}$$ or $$\sin54^{\circ}-\sin18^{\circ}=\frac{1}{2},$$ which is true because $$\sin54^{\circ}-\sin18^{\circ}=2\sin18^{\circ}\cos36^{\circ}=2\cos36^{\circ}\cos72^{\circ}=$$ $$=\frac{4\sin36^{\circ}\cos36^{\circ}\cos72^{\circ}}{2\sin36^{\circ}}=\frac{2\sin72^{\circ}\cos72^{\circ}}{2\sin36^{\circ}}=\frac{\sin144^{\circ}}{2\sin36^{\circ}}=\frac{1}{2}.$$

Answer 2

Hint: see that $$\cos(6)=\cos(30-24)=\cos(30)\cos(24)+\sin(30)\sin(24) =\frac{\sqrt{3}}2\cos(24) +\frac{1}2\sin(24)$$