How do you evaluate $ \cos \left( {\dfrac{{11\pi }}{6}} \right)? $

Hint: To evaluate the value of the given trigonometric function $ \cos t $ , first find the value for $ \cos 2t $ and then use the trigonometric identities of compound angle formula for cosine function and also find the quadrant in which the given angle lies and put the sign according to the quadrant.
Cosine has positive values in the first and fourth quadrant and negative in the second and third quadrant.
Also $ \cos (t + 2n\pi ) = \cos t\;{\text{and}}\;\cos ( - t) = \cos t $ , where “n” is integer.
The following formula will be used
 $ \cos 2t = 2{\cos ^2}t - 1 $

Complete step-by-step answer:
In order to evaluate the value of $ \cos \left( {\dfrac{{11\pi }}{6}} \right) $ we will first find the value of $ \cos 2 \times \left( {\dfrac{{11\pi }}{6}} \right) = \cos \left( {\dfrac{{22\pi }}{6}} \right) $
And we can write $ \cos \left( {\dfrac{{22\pi }}{6}} \right) $ as follows
 $ \Rightarrow \cos \left( {\dfrac{{22\pi }}{6}} \right) = \cos \left( {\dfrac{{24\pi }}{6} - \dfrac{{2\pi }}{6}} \right) $
And can further write it as
 $
   \Rightarrow \cos \left( {\dfrac{{22\pi }}{6}} \right) = \cos \left( {4\pi - \dfrac{\pi }{3}} \right) \\
   \Rightarrow \cos \left( {\dfrac{{22\pi }}{6}} \right) = \cos \left( {2 \times 2\pi - \dfrac{\pi }{3}} \right) \\
  $
From the periodic property of cosine, we can write $ \cos (t + 2n\pi ) = \cos t $ , where “n” is an integer
 $
   \Rightarrow \cos \left( {\dfrac{{22\pi }}{6}} \right) = \cos \left( {2 \times 2\pi - \dfrac{\pi }{3}} \right) \\
   \Rightarrow \cos \left( {\dfrac{{22\pi }}{6}} \right) = \cos \left( { - \dfrac{\pi }{3}} \right) \;
  $
Now we know that, $ \cos ( - t) = \cos t $
 $ \therefore \cos \left( {\dfrac{{22\pi }}{6}} \right) = \cos \left( {\dfrac{\pi }{3}} \right) = \dfrac{1}{2} $
We get the value for $ \cos \left( {\dfrac{{22\pi }}{6}} \right) = \cos \left( {\dfrac{{2 \times 11\pi }}{6}} \right) = \dfrac{1}{2} $
Now from the compound angle formula of cosine function, we know that
 $ \cos 2t = 2{\cos ^2}t - 1 $
So using this formula to evaluate the value of $ \cos \left( {\dfrac{{11\pi }}{6}} \right) $
 $
   \Rightarrow \cos \left( {\dfrac{{2 \times 11\pi }}{6}} \right) = 2{\cos ^2}\left( {\dfrac{{11\pi }}{6}} \right) - 1 \\
   \Rightarrow \dfrac{1}{2} + 1 = 2{\cos ^2}\left( {\dfrac{{11\pi }}{6}} \right) \\
   \Rightarrow \dfrac{3}{2} = 2{\cos ^2}\left( {\dfrac{{11\pi }}{6}} \right) \\
   \Rightarrow 2{\cos ^2}\left( {\dfrac{{11\pi }}{6}} \right) = \dfrac{3}{2} \\
   \Rightarrow {\cos ^2}\left( {\dfrac{{11\pi }}{6}} \right) = \dfrac{3}{4} \\
   \Rightarrow \cos \left( {\dfrac{{11\pi }}{6}} \right) = \pm \sqrt {\dfrac{3}{4}} \\
   \Rightarrow \cos \left( {\dfrac{{11\pi }}{6}} \right) = \dfrac{{ \pm \sqrt 3 }}{2} \;
  $
Since the angle $ \left( {\dfrac{{11\pi }}{6}} \right) $ is in the range of fourth quadrant i.e. $ \left[ {\dfrac{{3\pi }}{2},\;2\pi } \right] $
And in the fourth quadrant cosine function has always positive value. Therefore the negative result will be neglected in this question.
Therefore $ \dfrac{{\sqrt 3 }}{2} $ is the required value of $ \cos \left( {\dfrac{{11\pi }}{6}} \right) $
So, the correct answer is “ $ \dfrac{{\sqrt 3 }}{2} $ ”.

Note: When solving this type of trigonometric questions, where the argument or angle is greater than the principal argument of the trigonometric function then divide the argument by $ 2\pi $ and leave the quotient and take the remainder as the new argument, this process will make the problem easy to solve by shrinking the given argument into the range of principal argument.