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Question: How do you evaluate \({\cos ^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{2}} \right)\)?...

How do you evaluate cos⁡−1(−32){\cos ^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{2}} \right)?

Explanation

Solution

In order to find the solution of a trigonometric equation, we start by taking the inverse trigonometric function like inverse sin, inverse cosine, inverse tangent on both sides of the equation and then set up reference angles to find the rest of the answers.
For sin⁡−1{\sin ^{ - 1}} function, the principal value branch is [−π2,π2]\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right].
For cos⁡−1{\cos ^{ - 1}} function, the principal value branch is [0,π]\left[ {0,\pi } \right].
For tan⁡−1{\tan ^{ - 1}} function, the principal value branch is (−π2,π2)\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right).

Complete step by step solution:
According to definition of inverse ratio,
Ifcos⁡x=−32\cos x = - \dfrac{{\sqrt 3 }}{2},
Then, cos⁡−1(−32)=x{\cos ^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{2}} \right) = x where the value of x lies in the range [0,π]\left[ {0,\pi } \right].
Now, we know that the cosine function is positive in the first and fourth quadrants and negative in the second and third quadrant.
So, the angle x=cos⁡−1(−32)x = {\cos ^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{2}} \right) must lie either in second quadrant or in third quadrant.
We know that the value of cos⁡(π6)\cos \left( {\dfrac{\pi }{6}} \right) is (32)\left( {\dfrac{{\sqrt 3 }}{2}} \right). Also, cos⁡(π−θ)=−cos⁡(θ)\cos \left( {\pi - \theta } \right) = - \cos \left( \theta \right).
So, cos⁡(π−π6)=−cos⁡(π6)\cos \left( {\pi - \dfrac{\pi }{6}} \right) = - \cos \left( {\dfrac{\pi }{6}} \right) .
Therefore, cos⁡(5π6)=(−32)\cos \left( {\dfrac{{5\pi }}{6}} \right) = \left( { - \dfrac{{\sqrt 3 }}{2}} \right).
Hence, the value of cos⁡−1(−32){\cos ^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{2}} \right) is (5π6)\left( {\dfrac{{5\pi }}{6}} \right).

Note: The basic inverse trigonometric functions are used to find the missing angles in right triangles. While the regular trigonometric functions are used to determine the missing sides of the right-angled triangles, using the following formulae:
sin⁡θ=(Opposite SideHypotenuse)\sin \theta = \left( {\dfrac{{{\text{Opposite Side}}}}{{{\text{Hypotenuse}}}}} \right)
cos⁡θ=(Adjacent SideHypotenuse)\cos \theta = \left( {\dfrac{{{\text{Adjacent Side}}}}{{{\text{Hypotenuse}}}}} \right)
tan⁡θ=(Opposite SideAdjacent Side)\tan \theta = \left( {\dfrac{{{\text{Opposite Side}}}}{{{\text{Adjacent Side}}}}} \right)
Besides the trigonometric functions and inverse trigonometric functions, we also have some rules related to trigonometry such as the sine rule and cosine rule. According to the sine rule, the ratio of the sine of two angles is equal to the ratio of the lengths of the sides of the triangle opposite to both the angles. So, (sin⁡Asin⁡B)=(ab)\left( {\dfrac{{\sin A}}{{\sin B}}} \right) = \left( {\dfrac{a}{b}} \right).