
Find the principal value of : \[{\cos ^{ - 1}}( - 1)\].
Answer
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Hint:We have to find the principal value of the given trigonometric function \[{\cos ^{ - 1}}( - 1)\] . We solve this question using the concept of range of the inverse trigonometric functions . We should also have the knowledge of the values of the trigonometric function for different values of angles . We should also know how to use the inverse of trigonometric functions . We will change the given value into its value related to the given trigonometric function such that the value of angle of the covered value should lie in the range of the given trigonometric function and then solving the expression of inverse we will get the principal value of the given trigonometric function.
Complete step by step answer:
Given: Principal value of : \[{\cos ^{ - 1}}( - 1)\]
Since , the given trigonometric function is a cosine function . So , we would convert the value inside the given function in terms of angles of the cosine function so , as the inverse cancels out and we get the principal value of the given function . We would convert the value such that it would lie in the range of the cosine function.We know that the range of inverse of cosine function is given as :
\[{\cos ^{ - 1}}(\cos x) = x\] ; where \[x \in [0,\pi ]\]
We also know that the value of cosine function is \[ - 1\] for :
\[\cos \pi = - 1\]
Substituting the value , we get
\[{\cos ^{ - 1}}( - 1) = {\cos ^{ - 1}}(\cos \pi )\]
As the value of angle lies in the range , so using the concept of range we can write the expression as :
\[{\cos ^{ - 1}}( - 1) = \pi \]
Hence , the principal value of the given trigonometric function \[{\cos ^{ - 1}}( - 1)\] is \[\pi \].
Note:We used the concept of the range of the inverse of trigonometric functions . The range of the trigonometric function is stated as the value of the intervals between which all the values of the trigonometric function lie.Various inverse trigonometric functions are given as :
\[{\sin ^{ - 1}}( - x) = - {\sin ^{ - 1}}x\] , \[x \in [ - 1,1]\]
\[\Rightarrow {\cos ^{ - 1}}( - x) = \pi - {\cos ^{ - 1}}x\] , \[x \in [ - 1,1]\]
\[\Rightarrow {\tan ^{ - 1}}( - x) = - {\tan ^{ - 1}}x\] , \[x \in R\]
Complete step by step answer:
Given: Principal value of : \[{\cos ^{ - 1}}( - 1)\]
Since , the given trigonometric function is a cosine function . So , we would convert the value inside the given function in terms of angles of the cosine function so , as the inverse cancels out and we get the principal value of the given function . We would convert the value such that it would lie in the range of the cosine function.We know that the range of inverse of cosine function is given as :
\[{\cos ^{ - 1}}(\cos x) = x\] ; where \[x \in [0,\pi ]\]
We also know that the value of cosine function is \[ - 1\] for :
\[\cos \pi = - 1\]
Substituting the value , we get
\[{\cos ^{ - 1}}( - 1) = {\cos ^{ - 1}}(\cos \pi )\]
As the value of angle lies in the range , so using the concept of range we can write the expression as :
\[{\cos ^{ - 1}}( - 1) = \pi \]
Hence , the principal value of the given trigonometric function \[{\cos ^{ - 1}}( - 1)\] is \[\pi \].
Note:We used the concept of the range of the inverse of trigonometric functions . The range of the trigonometric function is stated as the value of the intervals between which all the values of the trigonometric function lie.Various inverse trigonometric functions are given as :
\[{\sin ^{ - 1}}( - x) = - {\sin ^{ - 1}}x\] , \[x \in [ - 1,1]\]
\[\Rightarrow {\cos ^{ - 1}}( - x) = \pi - {\cos ^{ - 1}}x\] , \[x \in [ - 1,1]\]
\[\Rightarrow {\tan ^{ - 1}}( - x) = - {\tan ^{ - 1}}x\] , \[x \in R\]
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