Question

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

Answer 1

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 $\dfrac{5\pi }{6}$

& \therefore y=\dfrac{5\pi }{6} \\\ & \therefore y={{\cos }^{-1}}\left[ \cos \left( \dfrac{7\pi }{6} \right) \right]=\dfrac{5\pi }{6} \\\ \end{aligned}$$ **Note:** We can use this question to make a general property, as the angle $$\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{aligned} & \Rightarrow \pi +\theta =\dfrac{7\pi }{6} \\\ & \Rightarrow \pi +\theta -\pi =\dfrac{7\pi }{6}-\pi \\\ & \therefore \theta =\dfrac{\pi }{6} \\\ \end{aligned}$$ Hence, the value of $${{\cos }^{-1}}\left[ \cos \left( x \right) \right]$$ equals $$\pi -\theta $$, $$x$$ is a third quadrant angle, of the form $$\pi +\theta $$.