y=-cos2(x-pi/12)
= -sin2(x-pi/12+3pi/12)
= -sin2(x+pi/6)
I don't understand how you get the 3pi/12. If there is another simpler way to get the same answer, please include it.=]
This uses the identity: $\displaystyle \cos \theta = \sin \left(\theta + {\color{blue}\: \frac{\pi}{2}}\right)$
So looking at your expression: $\displaystyle -\cos \left[2 \left(x - \frac{\pi}{12}\right)\right]$
$\displaystyle = - \sin \left[2 \left(x - \frac{\pi}{12}\right){\color{blue} \: + \: \frac{\pi}{2}} \: \right]$
$\displaystyle = - \sin \left[2 \left(x - \frac{\pi}{12}\right){\color{blue} \: + \: 2\left( \frac{\pi}{4}\right)} \: \right]$
$\displaystyle = - \sin \left[2 \left(x - \frac{\pi}{12}{\color{blue} \: + \: \frac{\pi}{4}}\right) \: \right]$ (Factored out a 2 of the inside expression)
etc. etc.