Question

Find the principal value of ${{\sin }^{-1}}\left( \sin 100 \right)$.

Answer 1

Hint: In order to solve this question, we require some basic knowledge of the concept of the principal value that is, for ${{\sin }^{-1}}x$, if $\theta $ is the principal value of ${{\sin }^{-1}}x$, then $0\le \theta \le \dfrac{\pi }{2}$. Now, here we will convert ${{\sin }^{-1}}\left( \sin 100 \right)$ as ${{\sin }^{-1}}x$, then we will find the principal value of ${{\sin }^{-1}}\left( \sin 100 \right)$.
Complete step-by-step answer:
In this question, we have been asked to find the principal value of ${{\sin }^{-1}}\left( \sin 100 \right)$. For that, we will consider $\sin 100=x$. Now, we know that 100 = 180 – 80. So, we can write, $\sin 100=\sin \left( 180-80 \right)=x$. Now, we know that $\sin \left( 180-\theta \right)=\sin \theta $. So, we can write, $\sin \left( 180-\theta \right)=\sin 80=x$.
$\sin 100=\sin 80=x.........(i)$
So, we can write ${{\sin }^{-1}}\left( \sin 100 \right)$ as ${{\sin }^{-1}}x$ or ${{\sin }^{-1}}\left( \sin 100 \right)={{\sin }^{-1}}x.........(ii)$.
We also know that equation (i) can also be written as $80={{\sin }^{-1}}x.........(iii)$.
Now, from equation (iii) we will substitute the value of ${{\sin }^{-1}}x$ to equation (ii). So, we will get as follows.
${{\sin }^{-1}}\left( \sin 100 \right)=80$
Now, we know that if $\theta $ is the principal value of ${{\sin }^{-1}}x$, then $0\le \theta \le \dfrac{\pi }{2}$. So, we will check whether 80 is the principal value or not by checking whether 80 lies between $\left[ 0,\dfrac{\pi }{2} \right]$. We know that if ${{1}^{\circ }}=\dfrac{\pi }{180}$ radian, we can say that ${{90}^{\circ }}=\dfrac{\pi }{2}$ radian. Therefore, we can write the range as $0\le \theta \le 90$.
And we know that general solution of $\theta ={{\sin }^{-1}}x$ is $\theta =n\left( 180 \right)+{{\left( -1 \right)}^{n}}{{\sin }^{-1}}x$. Therefore, we can say that the general solution of $\theta =n\left( 180 \right)+{{\left( -1 \right)}^{n}}80$. So, to get the principal, we will put a value of n as 0. Therefore, we get $\theta =80$. So, in this question, we get the principal value of ${{\sin }^{-1}}\left( \sin 100 \right)$ as 80˚ and we know that ${{0}^{\circ }}\le {{80}^{\circ }}\le {{90}^{\circ }}$. Hence, we can say that 80˚ is the principal value of ${{\sin }^{-1}}\left( \sin 100 \right)$.

Note: While solving this question, the possible mistake that the students can make is by not converting 100 to 180 – 80 and this will give them the wrong answers. This is because 100 does not lie between $\left[ 0,\dfrac{\pi }{2} \right]$. Also, we need to remember that $\sin \left( 180-\theta \right)=\sin \theta $, so that by using this identity we will be able to write $\sin 100=\sin \left( 180-80 \right)=\sin 80$.