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 $\cos(\pi+x)$?

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.
