Question

How do you evaluate \[{{\cos }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right)\]?

Answer 1

Hint: The inverse trigonometric functions give the value of an angle that lies in their respective principal range. The principal range for all inverse trigonometric functions is different.
For \[{{\sin }^{-1}}x\] it is \[\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]\], for \[{{\cos }^{-1}}x\] it is \[\left[ 0,\pi \right]\], for \[{{\tan }^{-1}}x\] it is \[\left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)\], for \[{{\cot }^{-1}}x\] it is \[\left( 0,\pi \right)\], for \[{{\sec }^{-1}}x\] it is \[\left[ 0,\dfrac{\pi }{2} \right)\bigcup \left( \dfrac{\pi }{2},\pi \right]\], and for \[{{\csc }^{-1}}x\] it is \[\left[ -\dfrac{\pi }{2},0 \right)\bigcup \left( 0,\dfrac{\pi }{2} \right]\].

Complete answer:
We are asked to find the value of \[{{\cos }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right)\]. We know that the inverse trigonometric functions \[{{T}^{-1}}\left( x \right)\], where \[T\] is a trigonometric function gives the value of an angle that lies in their respective principal range. The principal range for the inverse trigonometric function \[{{\cos }^{-1}}\left( x \right)\] is \[\left[ 0,\pi \right]\].
We have to find the value of \[{{\cos }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right)\], which means here \[x=\dfrac{1}{\sqrt{2}}\]. Let’s assume the value of \[{{\cos }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right)\] is \[y\]. \[y\] is an angle in the principal range of \[{{\cos }^{-1}}\left( x \right)\].
Hence, \[{{\cos }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right)=y\]
Taking \[\cos \] of both sides of the above equation we get,
\[\Rightarrow \cos \left( {{\cos }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right) \right)=\cos \left( y \right)\]
We know that \[T\left( {{T}^{-1}}\left( x \right) \right)=x\], \[T\] is a trigonometric function. Using this property in the above equation we get,
\[\Rightarrow \cos \left( y \right)=\dfrac{1}{\sqrt{2}}\]
As we know y is an angle in the range of \[\left[ 0,\pi \right]\]. Whose cosine gives \[\dfrac{1}{\sqrt{2}}\]. In the range of \[\left[ 0,\pi \right]\] there is only one such angle whose cosine gives \[\dfrac{1}{\sqrt{2}}\]. It is \[\dfrac{\pi }{4}\]. Hence, \[y=\dfrac{\pi }{4}\] radians.
So, the value of \[{{\cos }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right)\] is \[\dfrac{\pi }{4}\] radians.

Note: Generally inverse trigonometric functions will be asked for only those values, for which the angle can be found easily, so the values of the trigonometric functions of special angles should be remembered. The principal range of all inverse trigonometric functions should also be remembered.