Question

How do you find the value of \[\cos \left( {\dfrac{{7\pi }}{6}} \right)\]?

Answer 1

Hint:
In the given question, we have been asked to find the value of a trigonometric function. Now, the argument of the given trigonometric function is not in the range of the known values of the trigonometric functions as given in the standard table, in which values lie from \[0\] to \[\pi /2\]. But we can calculate that by using the formula of periodicity of the given trigonometric function and then solving it.

Formula Used:
We are going to use the formula of negative periodicity of cosine function, which is:
\[\cos \left( {\pi + \theta } \right) = - \cos \left( \theta \right)\]

Complete step by step solution:
Here, we have to calculate the value of \[\cos \left( {\dfrac{{7\pi }}{6}} \right)\].
Now, we know that the negative periodicity of the cosine function is \[\pi \], i.e., after each \[\pi \] difference, the value of the cosine function changes but only in the sign (positive, negative).
Hence, \[\cos \left( {\dfrac{{7\pi }}{6}} \right) = \cos \left( {\pi + \dfrac{\pi }{6}} \right) = - \cos \left( {\dfrac{\pi }{6}} \right)\]
Now, we know that
\[\cos \left( {\dfrac{\pi }{6}} \right) = \dfrac{{\sqrt 3 }}{2}\]

Hence, \[\cos \left( {\dfrac{{7\pi }}{6}} \right) = - \dfrac{{\sqrt 3 }}{2}\]

Additional Information:
In the given question, we applied the concept of negative periodicity of the cosine function, so it is necessary that we know the periodicity of each trigonometric function. Negative periodicity of sine, cosine, cosecant and secant is \[\pi \] while the periodicity of tangent and cotangent is \[\pi /2\].

Note:
We just need to remember the periodicity of the trigonometric functions, like here, the cosine function has the negative periodicity of \[\pi \], i.e., every \[\cos \] value repeats after this interval but changes its sign, so, when the angle is more than \[180^\circ \] or \[\pi \], we convert it to the smaller value and solve for the answer using the smaller value.