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

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

Explanation

Solution

Hint: In order to solve this question, we require some basic knowledge of the concept of the principal value that is, for sin⁡−1x{{\sin }^{-1}}x, if θ\theta is the principal value of sin⁡−1x{{\sin }^{-1}}x, then 0≤θ≤π20\le \theta \le \dfrac{\pi }{2}. Now, here we will convert sin⁡−1(sin⁡100){{\sin }^{-1}}\left( \sin 100 \right) as sin⁡−1x{{\sin }^{-1}}x, then we will find the principal value of sin⁡−1(sin⁡100){{\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(sin⁡100){{\sin }^{-1}}\left( \sin 100 \right). For that, we will consider sin⁡100=x\sin 100=x. Now, we know that 100 = 180 – 80. So, we can write, sin⁡100=sin⁡(180−80)=x\sin 100=\sin \left( 180-80 \right)=x. Now, we know that sin⁡(180−θ)=sin⁡θ\sin \left( 180-\theta \right)=\sin \theta . So, we can write, sin⁡(180−θ)=sin⁡80=x\sin \left( 180-\theta \right)=\sin 80=x.
sin⁡100=sin⁡80=x.........(i)\sin 100=\sin 80=x.........(i)
So, we can write sin⁡−1(sin⁡100){{\sin }^{-1}}\left( \sin 100 \right) as sin⁡−1x{{\sin }^{-1}}x or sin⁡−1(sin⁡100)=sin⁡−1x.........(ii){{\sin }^{-1}}\left( \sin 100 \right)={{\sin }^{-1}}x.........(ii).
We also know that equation (i) can also be written as 80=sin⁡−1x.........(iii)80={{\sin }^{-1}}x.........(iii).
Now, from equation (iii) we will substitute the value of sin⁡−1x{{\sin }^{-1}}x to equation (ii). So, we will get as follows.
sin⁡−1(sin⁡100)=80{{\sin }^{-1}}\left( \sin 100 \right)=80
Now, we know that if θ\theta is the principal value of sin⁡−1x{{\sin }^{-1}}x, then 0≤θ≤π20\le \theta \le \dfrac{\pi }{2}. So, we will check whether 80 is the principal value or not by checking whether 80 lies between [0,π2]\left[ 0,\dfrac{\pi }{2} \right]. We know that if 1∘=π180{{1}^{\circ }}=\dfrac{\pi }{180} radian, we can say that 90∘=π2{{90}^{\circ }}=\dfrac{\pi }{2} radian. Therefore, we can write the range as 0≤θ≤900\le \theta \le 90.
And we know that general solution of θ=sin⁡−1x\theta ={{\sin }^{-1}}x is θ=n(180)+(−1)nsin⁡−1x\theta =n\left( 180 \right)+{{\left( -1 \right)}^{n}}{{\sin }^{-1}}x. Therefore, we can say that the general solution of θ=n(180)+(−1)n80\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 θ=80\theta =80. So, in this question, we get the principal value of sin⁡−1(sin⁡100){{\sin }^{-1}}\left( \sin 100 \right) as 80˚ and we know that 0∘≤80∘≤90∘{{0}^{\circ }}\le {{80}^{\circ }}\le {{90}^{\circ }}. Hence, we can say that 80˚ is the principal value of sin⁡−1(sin⁡100){{\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 [0,π2]\left[ 0,\dfrac{\pi }{2} \right]. Also, we need to remember that sin⁡(180−θ)=sin⁡θ\sin \left( 180-\theta \right)=\sin \theta , so that by using this identity we will be able to write sin⁡100=sin⁡(180−80)=sin⁡80\sin 100=\sin \left( 180-80 \right)=\sin 80.