
Find the principal value of ${{\cos }^{-1}}\left( \cos 20 \right)$.
Answer
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Hint: In order to solve this question, we require some basic knowledge on the concept of the principal value that is, for ${{\cos }^{-1}}x$, if $\theta$ is the principal value of ${{\cos }^{-1}}x$, then $0\le \theta \le \pi$. So, here, we will convert ${{\cos }^{-1}}\left( \cos 20 \right)$ to ${{\cos }^{-1}}x$, then we will find the principal value of ${{\cos }^{-1}}\left( \cos 20 \right)$.
Complete step-by-step answer:
In this question, we have been asked to find the principal value of ${{\cos }^{-1}}\left( \cos 20 \right)$. For that, we will start with cos 20 of ${{\cos }^{-1}}\left( \cos 20 \right)$. So, let us consider $\cos 20=x.........(i)$. Therefore, we can write as,
${{\cos }^{-1}}\left( \cos 20 \right)={{\cos }^{-1}}x.........(ii)$
Now, we know that equation (i) can be further written as follows.
$20={{\cos }^{-1}}x.........(iii)$
Now, from equation (iii), we will put the value of ${{\cos }^{-1}}x$ in equation (ii). So, we will get, ${{\cos }^{-1}}\left( \cos 20 \right)=20$.
Now, we know that if $\theta$ is the principal value of some ${{\cos }^{-1}}x$ then, $0\le \theta \le \pi$. So, we will check whether 20 is the principal value or not by checking if 20 lies between $\left[ 0,\pi \right]$ or not.
We know that ${{1}^{\circ }}=\dfrac{\pi }{180}$ radian. So we can say that ${{180}^{\circ }}=\pi$ radian. Hence, we can write the range of $\theta ={{\cos }^{-1}}x$ as \[{{0}^{\circ }}\le {{\theta }^{\circ }}\le {{180}^{\circ }}\].
And we know that general solution of $\theta ={{\cos }^{-1}}x$ is $\theta =2n\left( 180 \right)\pm {{\cos }^{-1}}x$. Therefore, we can say that the general solution of $\theta =2n\left( 180 \right)\pm 20$. So, to get the principal, we will put a value of n as 0. Therefore, we get $\theta =20$. So, in this question, we get the principal value of ${{\cos }^{-1}}\left( \cos 20 \right)$ as 20˚. And we know that ${{0}^{\circ }}\le {{20}^{\circ }}\le {{180}^{\circ }}$. So, we can say that 20 is the principal value of ${{\cos }^{-1}}\left( \cos 20 \right)$.
Note: While solving this question, there is a possibility that the students may solve this question by finding the value of cos 20. This is also correct, but that will make the solution much more complicated and lengthier. So, the better method will be assuming $\cos 20=x$ and then solving the question.
Complete step-by-step answer:
In this question, we have been asked to find the principal value of ${{\cos }^{-1}}\left( \cos 20 \right)$. For that, we will start with cos 20 of ${{\cos }^{-1}}\left( \cos 20 \right)$. So, let us consider $\cos 20=x.........(i)$. Therefore, we can write as,
${{\cos }^{-1}}\left( \cos 20 \right)={{\cos }^{-1}}x.........(ii)$
Now, we know that equation (i) can be further written as follows.
$20={{\cos }^{-1}}x.........(iii)$
Now, from equation (iii), we will put the value of ${{\cos }^{-1}}x$ in equation (ii). So, we will get, ${{\cos }^{-1}}\left( \cos 20 \right)=20$.
Now, we know that if $\theta$ is the principal value of some ${{\cos }^{-1}}x$ then, $0\le \theta \le \pi$. So, we will check whether 20 is the principal value or not by checking if 20 lies between $\left[ 0,\pi \right]$ or not.
We know that ${{1}^{\circ }}=\dfrac{\pi }{180}$ radian. So we can say that ${{180}^{\circ }}=\pi$ radian. Hence, we can write the range of $\theta ={{\cos }^{-1}}x$ as \[{{0}^{\circ }}\le {{\theta }^{\circ }}\le {{180}^{\circ }}\].
And we know that general solution of $\theta ={{\cos }^{-1}}x$ is $\theta =2n\left( 180 \right)\pm {{\cos }^{-1}}x$. Therefore, we can say that the general solution of $\theta =2n\left( 180 \right)\pm 20$. So, to get the principal, we will put a value of n as 0. Therefore, we get $\theta =20$. So, in this question, we get the principal value of ${{\cos }^{-1}}\left( \cos 20 \right)$ as 20˚. And we know that ${{0}^{\circ }}\le {{20}^{\circ }}\le {{180}^{\circ }}$. So, we can say that 20 is the principal value of ${{\cos }^{-1}}\left( \cos 20 \right)$.
Note: While solving this question, there is a possibility that the students may solve this question by finding the value of cos 20. This is also correct, but that will make the solution much more complicated and lengthier. So, the better method will be assuming $\cos 20=x$ and then solving the question.
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