Question

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

Answer 1

Hint: In order to solve this question, we need to know few of the trigonometric ratios like, $\cos 60{}^\circ =\dfrac{1}{2}$ and $\operatorname{cosec}90{}^\circ =1$, We also need to know that $\cos \left( \pi -\theta \right)=-\cos \theta$ and $\operatorname{cosec}\left( -\theta \right)=-\operatorname{cosec}\theta$. By using these properties, we can find the principal value of ${{\operatorname{cosec}}^{-1}}\left( 2\cos \dfrac{2\pi }{3} \right)$.
Complete step-by-step answer:
In this question, we are asked to find the principal value of ${{\operatorname{cosec}}^{-1}}\left( 2\cos \dfrac{2\pi }{3} \right)$. For that, we will first start from $\cos \dfrac{2\pi }{3}$. So, we will convert $\dfrac{2\pi }{3}$ from radian form to degree form. We know that, $\pi \text{ }radian=180{}^\circ$. So, we can write, $\dfrac{2\pi }{3}$ radian as,
$\begin{aligned} & =\dfrac{2}{3}\times 180{}^\circ \\\ & \Rightarrow \dfrac{2\pi }{3}\text{ }radian=120{}^\circ \\\ \end{aligned}$
Hence, we can write $\cos \dfrac{2\pi }{3}=\cos 120{}^\circ$. Now, we know that $120{}^\circ =180{}^\circ -60{}^\circ$. So, we can write $\cos 120{}^\circ =\cos \left( 180{}^\circ -60{}^\circ \right)$. And we also know that $\cos \left( 180{}^\circ -\theta \right)=-\cos \theta$. So, we can write, $\cos \left( 180{}^\circ -60{}^\circ \right)=-\cos 60{}^\circ$. Now, we know that $\cos 60{}^\circ =\dfrac{1}{2}$, so we can write $-\cos 60{}^\circ =-\dfrac{1}{2}$. Therefore, we can write,
$\cos \dfrac{2\pi }{3}=-\dfrac{1}{2}$
Now, we will multiply the above equation by 2. So, we get,
$\begin{aligned} & 2\cos \dfrac{2\pi }{3}=2\times \left( -\dfrac{1}{2} \right) \\\ & \Rightarrow 2\cos \dfrac{2\pi }{3}=-1 \\\ \end{aligned}$
So, we can say that we have to find ${{\operatorname{cosec}}^{-1}}\left( -1 \right)$ because $2\cos \dfrac{2\pi }{3}=-1$. Now, we know that $\operatorname{cosec}90{}^\circ =1$. So, we can write $-1=-\operatorname{cosec}90{}^\circ$. And we know that $\operatorname{cosec}\left( -\theta \right)=-\operatorname{cosec}\theta$, so we can write $-1=\operatorname{cosec}\left( -90{}^\circ \right)$. Now, we know that $\operatorname{cosec}\left( -90{}^\circ \right)=-1$ can also be written as $\left( -90{}^\circ \right)={{\operatorname{cosec}}^{-1}}\left( -1 \right)$. Hence, we can write ${{\operatorname{cosec}}^{-1}}\left( -1 \right)=-90{}^\circ$ and we know that $-1=2\cos \dfrac{2\pi }{3}$. So, we get, ${{\operatorname{cosec}}^{-1}}\left( 2\cos \dfrac{2\pi }{3} \right)=-90{}^\circ$.
So, if we talk about the range of principal value of ${{\operatorname{cosec}}^{-1}}x$, then it is \left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)-\left\\{ 0 \right\\}. And we know that the general solution of, $\theta ={{\operatorname{cosec}}^{-1}}x$ is $\theta =n\left( 180{}^\circ \right)+\left[ {{\left( -1 \right)}^{n}} \right]{{\operatorname{cosec}}^{-1}}x$. Therefore, we can say that the general solution of $\theta =n\left( 180{}^\circ \right)-\left[ {{\left( -1 \right)}^{n}} \right]\dfrac{\pi }{2}$. So, to get the principal, we will put the value of n as 0. Therefore, we get $\theta =-90{}^\circ$. And we know that $-90{}^\circ =-\dfrac{\pi }{2}$, because $180{}^\circ =\pi \text{ }radian$. So, $-90{}^\circ$ lies in the range.
Hence, we can say, $-90{}^\circ$ is the principal value of ${{\operatorname{cosec}}^{-1}}\left( 2\cos \dfrac{2\pi }{3} \right)$.
Note: While solving this question, we need to remember that the range of $\theta ={{\operatorname{cosec}}^{-1}}x$ for principal value is, \left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)-\left\\{ 0 \right\\}. And, so whatever value we obtain should lie in the range and if it does not lie, then we have to apply the transformation like $\cos \left( 180{}^\circ -\theta \right)=-\cos \theta$ and so on.