Question

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

Answer 1

Hint: sine, cosine and tangent are the main functions while cosecant, secant and cotangent are the reciprocal of sine, cosine and tangent respectively. Thus the given function can be converted in the form of sine easily. Signs of a trigonometric function are the same in the two of the four quadrants in the graph that’s why any trigonometric function can have many solutions. To solve this we need to know identities of supplementary angles.

Complete step-by-step answer:
Given, \[\cos \left[ {\dfrac{{ - 7\pi }}{6}} \right] \]
We know that \[\cos ( - \theta ) = \cos (\theta )\] ,
So above becomes,
 \[ \Rightarrow \cos \left[ {\dfrac{{ - 7\pi }}{6}} \right] = \cos \left[ {\dfrac{{7\pi }}{6}} \right] \]
But we can write \[\dfrac{{7\pi }}{6} = \pi + \dfrac{\pi }{6}\] ,
 \[ = \cos \left[ {\pi + \dfrac{\pi }{6}} \right] \]
But we know the supplementary angle identity \[\cos (\pi + \theta ) = - \cos \theta \] . The negative sign is because of the third quadrant and cosine is negative in the third quadrant.
 \[ = - \cos \left[ {\dfrac{\pi }{6}} \right] \]
We know \[\pi = {180^0}\] and \[ \Rightarrow \dfrac{\pi }{6} = \dfrac{{{{180}^0}}}{6} = {30^0}\] , then we have,
 \[ = - \cos ({30^0})\]
We know \[\cos ({30^0}) = \dfrac{{\sqrt 3 }}{2}\]
 \[ - \dfrac{{\sqrt 3 }}{2}\] .
Thus we have \[ \Rightarrow \cos \left[ {\dfrac{{ - 7\pi }}{6}} \right] = - \dfrac{{\sqrt 3 }}{2}\]
So, the correct answer is “$ - \dfrac{{\sqrt 3 }}{2}$”.

Note: A graph is divided into four quadrants, all the trigonometric functions are positive in the first quadrant, all the trigonometric functions are negative in the second quadrant except sine and cosine functions, tangent and cotangent are positive in the third quadrant while all others are negative and similarly all the trigonometric functions are negative in the fourth quadrant except cosine and secant. The trigonometric functions are periodic, that is their value repeats after a specific interval using this property we find their value in the other quadrants.