So when the value of alpha we ignore the negative sign on 3

When we express $4\cos x - 3\sin x$ in the form $R\cos(x + \alpha)$, we use the expansion:

$$R\cos(x + \alpha) = R(\cos x \cos \alpha - \sin x \sin \alpha)$$ $$R\cos(x + \alpha) = (R\cos \alpha)\cos x - (R\sin \alpha)\sin x$$

Now, we compare this to the given expression:

$$4\cos x - 3\sin x$$

By comparing the coefficients of $\cos x$:

$$R\cos \alpha = 4$$

By comparing the coefficients of $\sin x$:

$$-(R\sin \alpha) = -3$$

Multiplying both sides by $-1$, we get:

$$R\sin \alpha = 3$$

So, it's not that we ignored the negative sign on the 3. Instead, the negative sign in the original expression $4\cos x - 3\sin x$ matches the negative sign that naturally appears in the expansion of $R\cos(x + \alpha)$. This means that $R\sin \alpha$ itself is positive 3.

This is why when we calculate $\tan \alpha = \frac{R\sin \alpha}{R\cos \alpha} = \frac{3}{4}$, both the numerator and denominator are positive, placing $\alpha$ in the first quadrant ($0^\circ < \alpha < 90^\circ$).

Does that clarify it for you?