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Question: Principal value of\[{\cos ^{ - 1}}( - 1/2)\] is equal to A) \[\dfrac{{\rm{\pi }}}{{\rm{3}}}\] B)...

Principal value ofcos⁡−1(−1/2){\cos ^{ - 1}}( - 1/2) is equal to
A) π3\dfrac{{\rm{\pi }}}{{\rm{3}}}
B) 2π3\dfrac{{{\rm{2\pi }}}}{{\rm{3}}}
C) −π3\dfrac{{{\rm{ - \pi }}}}{{\rm{3}}}
D) −2π3\dfrac{{{\rm{ - 2\pi }}}}{{\rm{3}}}

Explanation

Solution

Here, we have to use the trigonometric functions to find out the value of the given equation. Firstly we have to put the value of the fraction present inside the cos inverse function in terms of cos. Then we have to remove its negative sign by using the properties of trigonometry. Then by simplification we get the value of the given function.

Complete step by step solution:
Given function is cos⁡−1(−1/2){\cos ^{ - 1}}( - 1/2)………… (1)
First step is to replace the value of the fraction present inside the bracket in terms of cos function.
As ⇒cos⁡(π3)=12\Rightarrow \cos \left( {\dfrac{{\rm{\pi }}}{{\rm{3}}}} \right) = \dfrac{1}{2}
So we substitute the value of the 12\dfrac{1}{2} in the equation (1), we get
⇒cos⁡−1(−cos⁡(π3))\Rightarrow {\cos ^{ - 1}}( - \cos \left( {\dfrac{{\rm{\pi }}}{{\rm{3}}}} \right)) …………… (2)
Now, we have to remote the negative sign present in the bracket. So using the trigonometric identity, cos⁡(π−θ)=−cos⁡θ\cos ({\rm{\pi }} - \theta ) = - \cos \theta
Therefore,⇒cos⁡(π−π3)=−cos⁡(π3)\Rightarrow \cos \left( {{\rm{\pi }} - \dfrac{{\rm{\pi }}}{{\rm{3}}}} \right) = - \cos \left( {\dfrac{{\rm{\pi }}}{{\rm{3}}}} \right)
Substitute the value of −cos⁡(π3) - \cos \left( {\dfrac{{\rm{\pi }}}{{\rm{3}}}} \right) in the equation (2), we get
⇒cos⁡−1(cos⁡(π−π3))\Rightarrow {\cos ^{ - 1}}\left( {\cos \left( {{\rm{\pi }} - \dfrac{{\rm{\pi }}}{{\rm{3}}}} \right)} \right)
Now, by simplification cos inverse will cut out by normal cos function, we get
⇒cos⁡−1(cos⁡(π−π3))=π−π3=3π−π3=2π3\Rightarrow {\cos ^{ - 1}}\left( {\cos \left( {{\rm{\pi }} - \dfrac{{\rm{\pi }}}{{\rm{3}}}} \right)} \right) = {\rm{\pi }} - \dfrac{{\rm{\pi }}}{{\rm{3}}} = \dfrac{{{\rm{3\pi - \pi }}}}{{\rm{3}}} = \dfrac{{{\rm{2\pi }}}}{{\rm{3}}}

Hence, value of the cos⁡−1(−1/2){\cos ^{ - 1}}( - 1/2)is equal to the 2π3\dfrac{{{\rm{2\pi }}}}{{\rm{3}}}
So, option B is correct.

Note:
Trigonometry is the study of the relationship between the angles and sides of a right triangle. Right Triangle is a triangle where one of its interior angles is a right angle (90 degrees). The relation between the sides and angles of a right triangle is the basis for trigonometry. The side opposite the right angle is called the hypotenuse. The sides adjacent to the right angle are called base. Generally Pythagoras theorem is used to determine the third side if two sides are given. Pythagoras theorem stated that in a right angled triangle the square of the long side is equal to the sum of the squares of the other two sides.
sin A = side opposite to angle Ahypotenuse{\text{sin A = }}\dfrac{{{\text{side opposite to angle A}}}}{{{\text{hypotenuse}}}}, cos A = side adjacent to angle Ahypotenuse{\text{cos A = }}\dfrac{{{\text{side adjacent to angle A}}}}{{{\text{hypotenuse}}}}, tan A = side opposite to angle Aside adjacent to angle A{\text{tan A = }}\dfrac{{{\text{side opposite to angle A}}}}{{{\text{side adjacent to angle A}}}}, cot A = 1tan A{\text{cot A = }}\dfrac{1}{{{\text{tan A}}}}, sec A = 1cos A{\text{sec A = }}\dfrac{1}{{{\text{cos A}}}}, cosec A = 1sin A{\text{cosec A = }}\dfrac{1}{{{\text{sin A}}}}