Use an appropriate identity to find the exact value of the expression. Cos(-7pi/12)
Using the identity $\cos(-x) = \cos(x)$, the exact value of $\cos\left(-\frac{7\pi}{12}\right)$ can be found as follows:
We know that $\frac{7\pi}{12}$ is in the third quadrant. In the third quadrant, the reference angle is $\frac{\pi}{12}$.
Using the special angle values, we have $\cos\left(-\frac{7\pi}{12}\right) = \cos\left(\pi - \frac{\pi}{12}\right) = -\cos\left(\frac{\pi}{12}\right)$.
The value of $\cos\frac{\pi}{12}$ is not a special angle value, so we will use the difference formula: $\cos(a - b) = \cos a \cos b + \sin a \sin b$.
Let $a = \frac{\pi}{6}$ and $b = \frac{\pi}{12}$.
Then, $\cos\left(\frac{\pi}{12}\right) = \cos\left(\frac{\pi}{6} - \frac{\pi}{6}\right) = \cos\left(\frac{\pi}{6}\right)\cos\left(\frac{\pi}{12}\right) + \sin\left(\frac{\pi}{6}\right)\sin\left(\frac{\pi}{12}\right)$.
Using the special values, $\cos\left(\frac{\pi}{12}\right) = \frac{\sqrt{6}+\sqrt{2}}{4}$ and $\sin\left(\frac{\pi}{12}\right) = \frac{\sqrt{6}-\sqrt{2}}{4}$.
Substituting these values, we have:
$\cos\left(\frac{\pi}{12}\right) = \left(\frac{\sqrt{6}+\sqrt{2}}{4}\right)\left(\frac{\sqrt{6}+\sqrt{2}}{4}\right) + \left(\frac{\sqrt{6}-\sqrt{2}}{4}\right)\left(\frac{\sqrt{6}-\sqrt{2}}{4}\right)$
$= \frac{6 + 2\sqrt{12} + 2 + 6 - 2\sqrt{12} + 2}{16}$
$= \frac{16}{16}$
$= 1$.
Therefore, $\cos\left(-\frac{7\pi}{12}\right) = -\cos\left(\frac{\pi}{12}\right) = -1$.