Given that $2\cos(x + 50) = \sin(x + 40)$ show that $\tan x = \frac{1}{3}\tan 40$ Given that:
$$
2\cos(x + 50) = \sin(x + 40)
$$
Show, without using a calculator, that:
$$
\tan x = \frac{1}{3}\tan 40
$$
I've got the majority of it:
$$
2\cos x\cos50-2\sin x\sin50=\sin x\cos40+\cos x\sin40\\
$$
$$
\frac{2\cos50 - \sin40}{2\sin50 + \cos40}=\tan x
$$
But then, checking the notes, it says to use $\cos50 = \sin40$ and $\cos40 =\sin50$; which I don't understand. Could somebody explain this final step?

Where you have left of  using $\cos(90^\circ-x)=\sin x,\sin(90^\circ-x)=\cos x$
$$\frac{2\cos50^\circ - \sin40^\circ}{2\sin50^\circ + \cos40^\circ}=\frac{2\sin40^\circ- \sin40^\circ}{2\cos40^\circ + \cos40^\circ}=?$$
