Question

Find the principal value of ${{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)$.

Answer 1

Hint: We will be using the concept of inverse trigonometric functions to solve the problem. We will first write $\dfrac{\sqrt{3}}{2}$ as $\cos \theta$ then we will use the fact that ${{\cos }^{-1}}\left( \cos x \right)=x$ for $x\in \left[ 0,\pi \right]$.

Complete step-by-step answer:

Now, we have to find the value of ${{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)$.
Now, we know that the value of $\cos \left( \dfrac{\pi }{6} \right)=\dfrac{\sqrt{3}}{2}.........\left( 1 \right)$
We have taken $\dfrac{\sqrt{3}}{2}=\cos \left( \dfrac{\pi }{6} \right)$ as in the view of the principal value convention, x is confined to be in $\left[ 0,\pi \right]$.
Now, we know that the graph of ${{\cos }^{-1}}\left( \cos x \right)$ is,

Now, we have to find the value of${{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)$.
We will use the equation (1) to substitute the value of $\dfrac{\sqrt{3}}{2}$. So, we have,
${{\cos }^{-1}}\left( \cos \left( \dfrac{\pi }{6} \right) \right)$
Also, we know that ${{\cos }^{-1}}\left( \cos x \right)=x$ for $x\in \left[ 0,\pi \right]$. So, we have,
${{\cos }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)=\dfrac{\pi }{6}$

Note: To solve these type of question it is important to note that we have used a fact that ${{\cos }^{-1}}\left( \cos x \right)=x$ only for$x\in \left[ 0,\pi \right]$. For another value of x the graph of ${{\cos }^{-1}}\left( \cos x \right)$ must be used to find the value.