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 $\cos(\pi+x)$?
 A: Recall your elementary trigonometric identity $\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$
So we may begin by expanding $\cos\left(\frac\pi2-x\right)$.$$\begin{align*}\cos\left(\frac\pi2-x\right)&=\cos\frac\pi2\cos(-x)-\sin\frac\pi2\sin(-x)\\&=0(\cos x)-1(-\sin x)\text{ knowing that }\cos x,\sin x\text{ are even and odd, respectively}\\&=\sin x\end{align*}$$
This can be seen graphically as the two sinusoidal waves are identical apart from a phase shift of $\frac\pi2$ rad:

Additionally, our functions are also periodic with period $2\pi$, i.e. for any integer $k$ we find that adding said multiple of $2\pi$ does not affect our result: $$\cos(x+2\pi k)=\cos x\\\sin(x+2\pi k)=\sin x$$
With this information, we may easily finish off our solution; recognize that by subtracting $2\pi$ from $\frac{3\pi}2-x$ yields our familiar $-\frac\pi2-x$. Recalling evenness of $\cos x$, i.e. $\cos(-x)=\cos x$, we rewrite $\cos\left(-\frac\pi2-x\right)=\cos\left(\frac\pi2+x\right)$. All we have left to do is recognize our phase shift.
A: HINT : as $\cos(2\pi+y)=\cos2\pi\cos y-\sin2\pi\sin y=\cos y$
As $$\left(\frac{3\pi}2-x\right)-\left(-\frac\pi2-x\right)=2\pi$$
$$\implies \left(\frac{3\pi}2-x\right)=2\pi+\left(-\frac\pi2-x\right)$$
A: Remember that cosine is $2\pi$ periodic this. This means that
$$\cos(x + 2n\pi) = \cos(x)$$
for any $x\in\mathbb{R}$ and for any integer $n\in\mathbb{Z}$. Basically, this says that if you can shift the graph of cosine left or right by $2\pi$ without changing it. Shifting it once gives you your desired equality since
$$\cos\left(-\frac{\pi}{2} -x\right)=\cos\left(2\pi - \frac{\pi}{2} - x\right)=\cos\left(\frac{3\pi}{2}-x\right)$$
Your second question is similar. Sine and cosine satisfy the shift equality
$$\sin\left(\frac{\pi}{2} + x\right)=\cos(x)$$
Graphically, this means that shifting the graph of sine by $\frac{\pi}{2}$ to the left gives the graph of cosine. Applying this property will give you your second equality.
A: 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 $\sin (\frac{3 \pi}{2} + x )  =  \cos( - [\pi + x ])   =  \cos( \pi + x )$  .
