<s>[INST] Symmetry properties of $\sin$ and $\cos$. Why does $\cos\left(\frac{3\pi}{2} - x\right) = \cos\left(-\frac{\pi}{2} - x\right)$? For a question such as: 

If $\sin(x) = 0.34$, find the value of $\cos\left(\frac{3\pi}{2} - x\right)$.

The solution says that:
\begin{align*}
\cos\left(\frac{3\pi}{2} - x\right) &= \cos\left(-\frac{\pi}{2} - x\right)\\
&= \cos\left(\frac{\pi}{2} + x\right)\\
&= -\sin(x)
\end{align*}
Why does $\cos\left(\frac{3\pi}{2} - x\right) = \cos\left(-\frac{\pi}{2} - x\right)$?
Similarly, if $\cos(x) = 0.6$, find $\sin\left(\frac{3\pi}{2} + x\right)$, the solution says $\sin\left(\frac{3\pi}{2} + x\right)$ is equal to $\cos(\pi+x)$. How did they get $\cos(\pi+x)$?
 [/INST] For your first question, recall that $-\frac{\pi}{2}$ is just a different "angle name" for the same direction as $\frac{3\pi}{2}$.  The first is obtained by turning clockwise from the positive x-axis ($\theta = 0$) by $90^{\circ}$ or $\frac{\pi}{2}$ radians, and the second by turning counter-clockwise from $\theta = 0$ by $270^{\circ}$ or $\frac{3\pi}{2}$ radians.
As for your second question, the solution is using what are called the "co-relations", which include
$$\sin \theta = \cos (\frac{\pi}{2} - \theta) , \cos \theta = \sin (\frac{\pi}{2} - \theta) ,  $$
$$\cos \theta = \sin (\frac{\pi}{2} + \theta) , -\sin \theta = \cos (\frac{\pi}{2} + \theta) , \text{etc.} $$
So we can write $\sin (\frac{3 \pi}{2} + x )  =  \cos(\frac{\pi}{2} - [\frac{3\pi}{2} + x ])  = \cos(-\pi - x)  =  \cos( - [\pi + x ])$ .  But we also know that  $\cos(-\theta) = \cos\theta$ , so $\sin (\frac{3 \pi}{2} + x )  =  \cos( - [\pi + x ])   =  \cos( \pi + x )$  .
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