The value of cos\((\frac{13\pi}6)\) is \(\frac{\sqrt3}2\)
Now,
∴ The question becomes cos–1\((\frac{\sqrt 3}2)\)

The range of principal value of cos–1 is [0,π] and cos\((\frac{\pi}6)=\frac{\sqrt3}2\)
Therefore, the value of cos–1(cos\((\frac{13\pi}6)\)) is \(\frac{\pi}6.\)