Prove that $ \tan40° + \sqrt 3 =4 \sin40° $ 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.
 A: 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? :)
A: Re-writing $\tan 40^{\circ} = \frac{\sin 40^{\circ}}{\cos 40^{\circ}}$ and multiplying through by $\cos 40^{\circ}$ gives you $$\sin 40^{\circ} + \sqrt{3}\cos 40^{\circ} = 4 \sin 40^{\circ}\cos 40^{\circ}$$
Dividing by $2$ and making use of the fact that $2\sin \theta \cos \theta = \sin 2\theta$ gives us $$\frac{1}{2}\sin 40^{\circ} + \frac{\sqrt{3}}{2}\cos 40^{\circ} = \sin 80^{\circ}$$
We can rewrite this as $$\cos 60^{\circ}\sin 40^{\circ} + \sin 60^{\circ}\cos 40^{\circ} = \sin 80^{\circ}$$
So we have, finally $$\sin 100^{\circ} = \sin 80^{\circ}$$
Which is trivially true, so what you have is true. 
A: Let's take $LHS$ as follows  $$LHS=\tan 40^\circ+\sqrt{3}=\tan 40^\circ+\tan 60^\circ$$ $$=\frac{\sin 40^\circ}{\cos 40^\circ}+\frac{\sin 60^\circ}{\cos 60^\circ}$$ $$=\frac{\sin 40^\circ\cos 60^\circ+\cos 40^\circ\sin 60^\circ}{\cos 40^\circ\cos 60^\circ}$$ $$=\frac{\sin (40^\circ+60^\circ)}{\cos 40^\circ\cos 60^\circ}$$ $$=\frac{\sin 2(50^\circ)}{\cos (90^\circ-50^\circ)\left(\frac{1}{2}\right)}$$ $$=\frac{2\times 2\sin 50^\circ \cos 50^\circ}{\sin 50^\circ}$$ $$=4\cos 50^\circ$$$$=\color{blue}{4\sin 40^\circ=RHS}$$
A: For $\cos y\ne0,$ $$2\cos3y\tan y+4\sin y$$
$$=\dfrac{2\cos3y\sin y+2(2\sin y\cos y)}{\cos y}$$
$$=\dfrac{\sin4y-\sin2y+2\sin2y}{\cos y}$$
$$=2\sin3y$$
Set $2\cos3y=-1,3y=360^\circ n\pm120^\circ$ where $n$ is any integer
$y=120^\circ n\pm40^\circ$ where $n=0,\pm1$
Here $n=0$
