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?

Recall the following:
(a) $Cos(x)=Sin(90-x)$;
(b) $Sin(x)=Cos(90-x)$;
(c) $Tan(x)=\frac{Sin(x)}{Cos(x)}$.
(a), (b) and (c) are standard trig formula, from where you can get that :$$Cos(50)^{\circ}=Sin(90-50)^{\circ}=Sin40^{\circ},$$
$$Cos(40)^{\circ}=Sin(90-40)^{\circ}=Sin50^{\circ},$$
$$Sin(50)^{\circ}=Cos(90-50)^{\circ}=Cos40^{\circ}$$
and $$Sin(40)^{\circ}=Cos(90-40)^{\circ}=Cos50^{\circ}.$$
You know that $2Cos(x+50)=Sin(x+40)$.
So apply either (a) or (b).
For instance, using (b) shows that $Sin(x+40)=Cos(90-(x+40))=Cos(50-x)$. Hence, $$2Cos(x+50)=Cos(50-x).$$
So, $$2Cos(x+50)-Cos(50-x)=0------(+)$$
You can now use the following:
$$Cos(A+B)=CosACosB-SinASinB-----(1)$$
$$Cos(A-B)=CosACosB+SinASinB------(2).$$
For instance, $$2Cos(x+50)=2CosxCos50-2SinxSin50$$ and $$Cos(50-x)=Cos50Cosx+Sin50Sinx$$.
Now, eqn (+) gives: $$CosxCos50-SinxSin50=0-----(++)$$
Recall that $Cos(A+B)=CosACosB-SinASinB$. So (++)  becomes:
$$Cos(x+50)^{\circ}=0.$$
As $Cos 90^{\circ}=0$, we know that $$x=40^{\circ}.$$
You can finish up from there!
