Question

Using the principal values, write the value of ${{\cos }^{-1}}\left( \dfrac{1}{2} \right)+2{{\sin }^{-1}}\left( \dfrac{1}{2} \right)$ .

Answer 1

Hint: In this question we are given to find the value of inverse trigonometric function using principal values. First we need to find the values of ${{\sin }^{-1}}\left( \dfrac{1}{2} \right)$ and ${{\cos }^{-1}}\left( \dfrac{1}{2} \right)$ then equate the expression to $'\theta '$ . The range of ${{\sin }^{-1}}$ lies between $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$ and ${{\cos }^{-1}}$ lies between the range of $\left[ 0,\pi \right]$ then simplify the values to get the answer.

Complete step-by-step answer:
A principal value of a function is the value selected at a point in the domain of a multiple-valued function, chosen so that the function has a single value at the point. The principal value of ${{\sin }^{-1}}x$for $x>0$ , is the length of the arc of a unit circle centered at the origin which subtends an angle at the centre whose sine is x.
The principal value of ${{\sin }^{-1}}$ lies between the range of $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$and ${{\cos }^{-1}}$ will lies between the range of $\left[ 0,\pi \right]$.
We have been given the functions ${{\cos }^{-1}}\left( \dfrac{1}{2} \right)+2{{\sin }^{-1}}\left( \dfrac{1}{2} \right)$ for which we need to find the principal value of ${{\cos }^{-1}}\left( \dfrac{1}{2} \right)$ and ${{\sin }^{-1}}\left( \dfrac{1}{2} \right)$.
To get the value of ${{\cos }^{-1}}\left( \dfrac{1}{2} \right)$ ,
Let us take the principal value of ${{\cos }^{-1}}\left( \dfrac{1}{2} \right)$ as $'\theta '$ .
Thus, ${{\cos }^{-1}}\left( \dfrac{1}{2} \right)=\theta$ .
We can write it as –
$\cos \theta =\left( \dfrac{1}{2} \right)$
We know that $\cos {{60}^{\circ }}=\dfrac{1}{2}$ .
$\theta ={{60}^{\circ }}=60\times \dfrac{\pi }{180}$
$=\dfrac{\pi }{3}$ .
$\therefore \theta ={{\cos }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{\pi }{3}$ ………………………. (1)
To find the value of ${{\sin }^{-1}}\left( \dfrac{1}{2} \right)$ ,
Let us take the principal value of ${{\sin }^{-1}}\left( \dfrac{1}{2} \right)$ as $'\theta '$ .
Thus, ${{\sin }^{-1}}\left( \dfrac{1}{2} \right)=\theta$ .
We can write it as –
$\sin \theta =\left( \dfrac{1}{2} \right)$
We know that $\sin {{30}^{\circ }}=\dfrac{1}{2}$ .
$\theta ={{30}^{\circ }}=30\times \dfrac{\pi }{180}$
$=\dfrac{\pi }{6}$ .
$\therefore \theta ={{\sin }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{\pi }{6}$………………………….. (2)
Now, let us find the value of ${{\cos }^{-1}}\left( \dfrac{1}{2} \right)+2{{\sin }^{-1}}\left( \dfrac{1}{2} \right)$ by substituting the value of ${{\cos }^{-1}}\left( \dfrac{1}{2} \right)$ and ${{\sin }^{-1}}\left( \dfrac{1}{2} \right)$ from equation (1) and (2).

& {{\cos }^{-1}}\left( \dfrac{1}{2} \right)+2{{\sin }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{\pi }{3}+2\times \dfrac{\pi }{6} \\\ & =\dfrac{\pi }{3}+\dfrac{\pi }{3} \\\ & =\dfrac{2\pi }{3} \\\ \end{aligned}$$ Hence, the value of $${{\cos }^{-1}}\left( \dfrac{1}{2} \right)+2{{\sin }^{-1}}\left( \dfrac{1}{2} \right)$$ using principal values is $\dfrac{2\pi }{3}$ . Note: Students should know what is the domain and range of trigonometric functions. The domain of a function is the specific set of values that the independent variable in a function can take on. The range is the resulting value that the dependent variable can have as x varies throughout the domain. They should not get confused between range of sine and cosine functions, as $${{\sin }^{-1}}$$ lies between the range of $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$ and ${{\cos }^{-1}}$ lies between the range of $\left[ 0,\pi \right]$.