Question

Write the principal value of $\left[ {{\cos }^{-1}}\dfrac{\sqrt{3}}{2}+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right) \right]$

Answer 1

Hint: To solve this type of problems we have to know the inverse trigonometric functions like ${{\cos }^{-1}}(\cos \theta )$ and ${{\sin }^{-1}}(\sin \theta )$. By applying formulas and writing the values of trigonometric functions we will get the values in the range of a function. Use ${{\cos }^{-1}}(-\theta )=\pi -{{\cos }^{-1}}\theta$.

Complete step-by-step answer:

Now writing the expression
$\left[ {{\cos }^{-1}}\dfrac{\sqrt{3}}{2}+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right) \right]$ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
We know that $\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}=\cos \dfrac{\pi }{6}$
$\cos {{60}^{\circ }}=\dfrac{1}{2}=\cos \dfrac{\pi }{3}$
${{\cos }^{-1}}(-\theta )=\pi -{{\cos }^{-1}}\theta$ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a)
Now rewriting the expression (1) we get
$\left[ {{\cos }^{-1}}\left( \cos \dfrac{\pi }{6} \right)+\pi -{{\cos }^{-1}}\left( \cos \dfrac{\pi }{3} \right) \right]$
We know that
${{\cos }^{-1}}(\cos \theta )=\theta \forall \theta \in \left[ 0,\pi \right]$. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (b)
Now from (b) further writing the expression we get,
= $\left[ \dfrac{\pi }{6}+\pi -\dfrac{\pi }{3} \right]$
= $\left[ \dfrac{5\pi }{6} \right]$

Note: The interval $\left[ 0,\pi \right]$ as shown in (b) is the range for the inverse trigonometric function ${{\cos }^{-1}}\theta$. ${{\cos }^{-1}}(\cos \theta )=\theta \forall \theta \in \left[ 0,\pi \right]$ this means the value becomes $\theta$ only in the given range. From (a) the range of ${{\cos }^{-1}}\theta$ is $\left[ 0,\pi \right]$ we have to use that expression of (a) because the range of ${{\cos }^{-1}}\theta$ is $\left[ 0,\pi \right]$. Take care while doing calculations.