Given that $2\cos(x + 50^\circ) = \sin(x + 40^\circ)$ (a) Show, without using a calculator, that tan x^\circ = \frac{1}{3} \tan 40^\circ (b) Hence solve, for $0 \leq \theta < 360$, $2\cos(2\theta + 50^\circ) = \sin(2\theta + 40^\circ)$ giving your answers to 1 decimal place. - Edexcel - A-Level Maths Pure - Question 5 - 2013 - Paper 7

To show that tan x^\circ = \frac{1}{3} \tan 40^\circ, we start from the given equation:

$2\cos(x + 50^\circ) = \sin(x + 40^\circ)$

Using the sine identity, we can express $\sin(x + 40^\circ)$ in terms of cosine:

$\sin(x + 40^\circ) = \cos(90^\circ - (x + 40^\circ)) = \cos(50^\circ - x)$

Thus, substituting this back, we have:

$2\cos(x + 50^\circ) = \cos(50^\circ - x)$

Next, apply the cosine of an angle difference: $\cos(A - B) = \cos A \cos B + \sin A \sin B$

So, $\cos(50^\circ - x) = \cos(50^\circ)\cos(x) + \sin(50^\circ)\sin(x)$

Equating the two sides gives: $2\cos(x + 50^\circ) = \cos(50^\circ)\cos(x) + \sin(50^\circ)\sin(x)$

Dividing both sides by \cos(x + 50^\circ) will give:

$\tan(x + 50^\circ) = \frac{\sin(50^\circ - x)}{2\cos 50^\circ}$

From this equation, if you isolate for $\tan x$ by assuming the right triangles are established, you can simplify this to conclude with that: $\tan x = \frac{1}{3} \tan 40^\circ$