Question

How do you calculate \[{{\cos }^{-1}}\left[ \cos \left( \dfrac{7\pi }{6} \right) \right]\]?

Answer 1

Hint: The given expression is of the form of the function \[f(g(x))\], here \[f(x)={{\cos }^{-1}}(x)\And g(x)=\cos (x)\]. To find the value of \[f(g(x))\] at \[x=a\], we first need to find the value of \[g(a)\], and then find the value of \[f(x)\] at \[x=g(a)\]. We will do the same for the given function.

Complete step by step answer:
We are asked to find the value of \[{{\cos }^{-1}}\left[ \cos \left( \dfrac{7\pi }{6} \right) \right]\]. Let there be a function \[{{\cos }^{-1}}\left( \cos x \right)\], we need to find the value of this function at \[x=\dfrac{7\pi }{6}\]. As this is the function of the form \[f(g(x))\], here \[f(x)={{\cos }^{-1}}(x)\And g(x)=\cos (x)\]. We first need to find the value of \[g(x)=\cos (x)\], at \[x=\dfrac{7\pi }{6}\].
\[\cos \left( \dfrac{7\pi }{6} \right)=\cos \left( \pi +\dfrac{\pi }{6} \right)\]
\[\Rightarrow -\cos \left( \dfrac{\pi }{6} \right)=-\dfrac{\sqrt{3}}{2}\]
Now that we have the value of \[g(x)=\cos (x)\], at \[x=\dfrac{7\pi }{6}\]. We have to find the value of \[f(x)={{\cos }^{-1}}(x)\], at \[x=\cos \left( \dfrac{7\pi }{6} \right)=-\dfrac{\sqrt{3}}{2}\]. By doing this we get
\[\Rightarrow {{\cos }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)\]
We know that the inverse trigonometric function gives the value of the angle in their principal range. The principal range of the inverse cosine function is \[\left[ 0,\pi \right]\].
We have to find the value of \[{{\cos }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)\], let the value of \[{{\cos }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)\] is \[y\], \[y\]is an angle in the principal range of cosine inverse function.
\[\Rightarrow y={{\cos }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right)\]
Taking \[\cos \] of both sides, we get
\[\Rightarrow \cos \left( y \right)=\cos \left( {{\cos }^{-1}}\left( -\dfrac{\sqrt{3}}{2} \right) \right)\]
\[\Rightarrow \cos \left( y \right)=-\dfrac{\sqrt{3}}{2}\]
We know that \[y\] is an angle in the range of \[\left[ 0,\pi \right]\], whose cosine gives the value \[-\dfrac{\sqrt{3}}{2}\]. There is only one such angle, and that is \[\dfrac{5\pi }{6}\]
\[\begin{align}
  & \therefore y=\dfrac{5\pi }{6} \\
 & \therefore y={{\cos }^{-1}}\left[ \cos \left( \dfrac{7\pi }{6} \right) \right]=\dfrac{5\pi }{6} \\
\end{align}\]

Note:
We can use this question to make a general property, as the angle \[\dfrac{7\pi }{6}\] is a third quadrant angle, it can be expressed as \[\pi +\theta \]. We can find the value of \[\theta \], by comparing it with the given angle.
\[\begin{align}
  & \Rightarrow \pi +\theta =\dfrac{7\pi }{6} \\
 & \Rightarrow \pi +\theta -\pi =\dfrac{7\pi }{6}-\pi \\
 & \therefore \theta =\dfrac{\pi }{6} \\
\end{align}\]
Hence, the value of \[{{\cos }^{-1}}\left[ \cos \left( x \right) \right]\] equals \[\pi -\theta \], \[x\] is a third quadrant angle, of the form \[\pi +\theta \].