在不使用计算器的情况下，化简下列三角函数。

1. Apply the addition formula for sine

$\sin \left(x + \frac{4\pi}{3}\right) = \sin x \cos \frac{4\pi}{3} + \cos x \sin \frac{4\pi}{3}$

2. Evaluate the exact values of the trigonometric functions for $4\pi/3$

$\cos \frac{4\pi}{3} = -\cos \frac{\pi}{3} = -\frac{1}{2}$
$\sin \frac{4\pi}{3} = -\sin \frac{\pi}{3} = -\frac{\sqrt{3}}{2}$

3. Substitute the exact values to simplify the expression

$\sin \left(x + \frac{4\pi}{3}\right) = \sin x \left(-\frac{1}{2}\right) + \cos x \left(-\frac{\sqrt{3}}{2}\right)$
$= -\frac{1}{2}\sin x - \frac{\sqrt{3}}{2}\cos x$

1. Express $75^\circ$ as the sum of two special angles and apply the cosine addition formula

$\cos 75^\circ = \cos(45^\circ + 30^\circ)$
$= \cos 45^\circ \cos 30^\circ - \sin 45^\circ \sin 30^\circ$

2. Substitute the exact values for the special angles and simplify

$= \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right)$
$= \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}$
$= \frac{\sqrt{6} - \sqrt{2}}{4}$

1. Apply the compound angle formula for tangent

$\tan(\theta - 45^\circ) = \frac{\tan \theta - \tan 45^\circ}{1 + \tan \theta \tan 45^\circ}$

2. Substitute the exact value of $\tan 45^\circ$ to simplify

Since $\tan 45^\circ = 1$,
$\frac{\tan \theta - 1}{1 + \tan \theta (1)} = \frac{\tan \theta - 1}{1 + \tan \theta}$