Trigonometric identity, possible error I need to prove the following trigonometric identity:
$$ \frac{\sin^2(\frac{5\pi}{6} - \alpha )}{\cos^2(\alpha - 4\pi)} - \cot^2(\alpha - 11\pi)\sin^2(-\alpha - \frac{13\pi}{2})  =\sin^2(\alpha)$$
I cannot express $\sin(\frac{5\pi}{6}-\alpha)$ as a function of $\alpha$. Could it be a textbook error?
 A: Since all the trig values are squared, it seems as though the exercise is simply playing with shifts by odd or even multiples of $\pi/2$.
Loosely,

  
*
  
*Shifting by "$\frac{\pi}{2} \cdot \text{odd}$" switches "sin" and "cos" (and possibly affects the sign)
  
*Shifting by "$\frac{\pi}{2} \cdot \text{even}$" ($=$ "$\pi \cdot \text{any}$") preserves "sin" and "cos" (and possibly affects sign)
  
*Negating the argument preserves "sin" and "cos" (and possibly affects sign)
  

Since squaring eliminates sign considerations, we have, simply:

$$\begin{align}
\mathrm{trig}^2\left( \pm \; \theta \pm \frac{\pi}{2} \text{odd} \right) &= \mathrm{cotrig}^2\theta \\
\mathrm{trig}^2\left( \pm \; \theta \pm \frac{\pi}{2} \text{even} \right) &= \mathrm{trig}^2\left( \pm \; \theta \pm \pi \cdot \text{any} \right) = \mathrm{trig}^2\theta
\end{align}$$

where each "$\pm$" is independent, "any" means (of course) "any integer", and "trig" can in fact be any of the six trig functions.
This makes pretty quick work of the simplification process ...
$$\begin{align}
\frac{\sin^2\left(\frac{5\pi}{6}-\alpha\right)}{\cos^2\left(\alpha-4\pi\right)} - \cot^2\left(\alpha-11\pi\right) \; \sin^2\left(-\alpha-\frac{13\pi}{2}\right) &\stackrel{?}{=} \sin^2\alpha \\[1em]
\frac{\sin^2\left(\frac{5\pi}{6}-\alpha\right)}{\cos^2\alpha} - \cot^2\alpha \; \cos^2\alpha &\stackrel{?}{=} \sin^2\alpha
\end{align}$$
... right up to the point at which the process shudders to a halt.
Given the nature of all the other terms (and @Adam's comment that sum and difference identities are not allowed), I suspect that "$\frac{5\pi}{6}$" is a typo of "$\frac{5\pi}{2}$", which would get us a little further ...
$$\frac{\cos^2\alpha}{\cos^2\alpha} - \cot^2\alpha \; \cos^2\alpha = 1 - \cot^2\alpha \;\cos^2\alpha \stackrel{?}{=} \sin^2\alpha$$
... but we hit another snag. Could it be that "$\sin^2\left(-\alpha-\frac{13\pi}{2}\right)$" is a typo of "$\cos^2(...)$"? If so, then that factor should've simplified to "$\sin^2\alpha$", and we'd have
$$1 - \cot^2\alpha \;\sin^2\alpha = 1 - \cos^2\alpha = \sin^2\alpha$$
as desired.
(It's also possible that, instead of a sin-cos typo, "$\cot$" is a typo for "$\tan$", but it seems like that would be an easier one for the OP to notice.)
A: Some important translations: 
$$\tag 1\sin(x\pm 2 \pi) = \sin x $$
$$\tag {1'}\cos(x\pm 2 \pi) = \cos x $$
$$\tag 2\cot(x\pm \pi)= \cot x$$
$$\tag {2'}\tan(x\pm \pi)= \tan x$$
$$\tag 3 \sin \left(\frac \pi 2 -x \right)=\cos x$$
$$\tag 4 \cos \left(\frac \pi 2 -x \right)=\sin x$$
$$\tag 5 \sin(\pi-x)=\sin x$$ and $$\tag 6\cos(\pi-x)=-\cos x$$
and $$\tag 7 \sin(-x)=-\sin x$$
$$\tag 8 \cos (-x) = \cos x$$ 
Although taking $\alpha =0$ reveals that the equality doesn't hold, assume that there is no typo, then, we could move on as follows.
You have that
$$ \frac{\sin^2 \left(\frac{5\pi}{6} - \alpha \right)}{\cos^2(\alpha - 4\pi)} - \cot^2(\alpha - 11\pi)\sin^2 \left(-\alpha - \frac{13\pi}{2}\right)  =\sin^2(\alpha)$$
Using the above, we can write
$$\eqalign{
  & {\sin ^2}\left( {\frac{{5\pi }}{6} - \alpha } \right) = {\left[ { - \sin \left( {\alpha  - \frac{{5\pi }}{6}} \right)} \right]^2} = {\sin ^2}\left( {\alpha  - \frac{{5\pi }}{6}} \right)  \cr 
  & {\cot ^2}(\alpha  - 11\pi ) = {\cot ^2}\left( {\alpha  - 10\pi } \right) =  \cdots  = {\cot ^2}\alpha   \cr 
  & {\sin ^2}\left( { - \alpha  - \frac{{13\pi }}{2}} \right) = {\left[ { - \sin \left( {\alpha  + \frac{{13\pi }}{2}} \right)} \right]^2} = \sin {\left( {\alpha  + \frac{{13\pi }}{2}} \right)^2}  \cr 
  & {\cos ^2}(\alpha  - 4\pi ) = {\cos ^2}(\alpha  - 2\pi ) = {\cos ^2}\alpha  \cr} $$
so that we have
$$\frac{{{{\sin }^2}\left( {\alpha  - \frac{{5\pi }}{6}} \right)}}{{{{\cos }^2}\alpha }} - {\cot ^2}\alpha {\sin ^2}\left( {\alpha  + \frac{{13\pi }}{2}} \right) = {\sin ^2}\alpha $$
Now
$${\sin ^2}\left( {\alpha  - \frac{{5\pi }}{6}} \right) = {\sin ^2}\left( {\alpha  - \frac{{3\pi }}{6} - \frac{{2\pi }}{6}} \right) = {\sin ^2}\left( {\alpha  - \frac{\pi }{3} - \frac{\pi }{2}} \right) = {\left( { - 1} \right)^2}{\sin ^2}\left( {\frac{\pi }{2} - \left( {\alpha  - \frac{\pi }{3}} \right)} \right) = {\cos ^2}\left( {\alpha  - \frac{\pi }{3}} \right)$$
and
$$\eqalign{
  & {\sin ^2}\left( {\alpha  + \frac{{13\pi }}{2}} \right) = {\sin ^2}\left( {\alpha  + \frac{{12\pi }}{2} + \frac{\pi }{2}} \right) = {\sin ^2}\left( {\alpha  + 6\pi  + \frac{\pi }{2}} \right) = {\sin ^2}\left( {\alpha  + \frac{\pi }{2}} \right)  \cr 
  &  = {\sin ^2}\left( {\frac{\pi }{2} - \left( { - \alpha } \right)} \right) = {\cos ^2}\left( { - \alpha } \right) = {\cos ^2}\alpha  \cr} $$
so that you have
$$\frac{{{{\cos }^2}\left( {\alpha  - \frac{\pi }{3}} \right)}}{{{{\cos }^2}\alpha }} - {\cot ^2}\alpha {\cos ^2}\alpha  = {\sin ^2}\alpha $$
Now, solving for $${{{\cos }^2}\left( {\alpha  - \frac{\pi }{3}} \right)}$$
gives
$${\cos ^2}\left( {\alpha  - \frac{\pi }{3}} \right) = {\cos ^2}\alpha {\sin ^2}\alpha  + {\cos ^6}\alpha \frac{1}{{{{\sin }^2}\alpha }}$$
There is some typo in your excercise, since letting $\alpha =0$ gives $1/4$ on the LHS and is not defined for the RHS.  When you discover what the typo is, move on with the listed translations. 
