Trigonometry – Prove that $ \tan40° + \sqrt 3 =4 \sin40° $

Question

The equality I'm trying to prove looks like that: $$ \tan40° + \sqrt 3 =4 \sin40° $$
My guess is that $\sqrt3$ can be rewritten as $\tan60°$ and I can use proved in previous exercise formula $$\tan3 \alpha = \frac{(3 – \tan^2\alpha)\tan\alpha}{1 – 3\tan\alpha}$$
But after dealing with it for an hour I feel misled. Some hints would be greatly appreciated.

Answer 1

Rewrite the equation using $\tan x = \frac{\sin x}{\cos x}$ $$ \frac{1}{2}\sin 40° + \frac{\sqrt{3}}{2}\cos 40° = 2 \sin 40° \cos 40° = \sin 80° = \cos 10° $$

Do you guess the formula on the left side? :)