Question

What is the principle value of \[{\cos ^{ - 1}}( - \dfrac{1}{2})\] ?

Answer 1

Hint: According to the question, principal values are the values along one chosen branch of that function, so that it is single-valued. Ranges of cosine function varies from \[ - 1\] to \[1\]. So, we can solve the question by taking the cosine function as an unknown value such as \[x\]or \[y\].

Formula used: \[\cos \left( {\pi - \theta } \right) = - \cos \theta \]and \[{\text{ }}cos \dfrac{\pi }{3} = \dfrac{1}{2}\]

Complete step-by-step answer:
The function from the question is \[{\cos ^{ - 1}}( - \dfrac{1}{2})\].
Let us assume that, \[{\cos ^{ - 1}}( - \dfrac{1}{2}) = x\]
Now, we can also write this as:
\[\cos x = \dfrac{1}{2}\]
\[\; \Rightarrow cosx = - {\text{ }}cos \dfrac{\pi }{3}\] (Because we know that the value of \[{\text{ }}cos \dfrac{\pi }{3} = \dfrac{1}{2}\]).
Now we can write:
\[ - {\text{ }}cos \dfrac{\pi }{3} = {\text{ }}cos{\text{ }}\left( {\pi - \dfrac{\pi }{3}} \right)\] (Because we know that \[\cos \left( {\pi - \theta } \right) = - \cos \theta \])
We know that the range of principal value branches of cosine function varies from \[0\,to\,\pi \].
So, value of \[\cos 2\dfrac{\pi }{3} = - \dfrac{1}{2}\]
Therefore, the principal value of \[{\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right)\] will be \[2\dfrac{\pi }{3}\].

Additional information: Value of cosine function is decreasing which decreases in the sequence \[1 > 0 > - 1\]in the interval \[{0^ \circ }\,to\,{180^ \circ }\].

Note: If we take an inverse trigonometric function, say \[\cos ( - 1)x\]for \[x > 0\]. Then the length of the arc of a unit circle which is centered at the origin, that subtends an angle at the center, whose cosine is \[x\], is the principal value of that inverse trigonometric function. We need to always remember the values of the trigonometric functions and their ranges also if we want to solve this type of problem.