<s>[INST] 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?
 [/INST] Using the well-known identity $$\cos (90^{\circ} - x) = \sin x$$ since the cosine function is just a $90^{\circ}$ horizontal translation of the sine function. 
Taking $x = 40^{\circ}$ gives us $\cos 50^{\circ} = \sin 40^{\circ}$ and letting $x=50^{\circ}$ establishes the second result. Then $$\begin{align}2\cos x\cos50-2\sin x\sin50=\sin x\cos40+\cos x\sin40 \\ \iff 2 \cos x \sin 40 - 2\sin x\cos 40 = \sin x \cos 40 +\cos x \sin 40 \\ \iff 2 \cos x\sin 40 - \cos x\sin 40 = \sin x\cos 40 + 2\sin x\cos 40 \\ \iff \tan 40 \cos x = 3 \sin x  \end{align}$$
so that $$\tan x = \frac{\tan 40^{\circ}}{3}$$
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