Question

How do you get the exact value of $ {{\csc }^{-1}}\left( 2 \right) $ ?

Answer 1

Hint: We convert the inverse function from $ {{\csc }^{-1}}\left( x \right) $ to $ {{\sin }^{-1}}\left( \dfrac{1}{x} \right) $ . Then we explain the function $ \arcsin \left( x \right) $ . We express the inverse function of sin in the form of $ \arcsin \left( x \right)={{\sin }^{-1}}x $ . We draw the graph of $ \arcsin \left( x \right) $ and the line $ x=\dfrac{1}{2} $ to find the intersection point as the solution.

Complete step-by-step answer:
First, we convert the inverse function from $ {{\csc }^{-1}}\left( x \right) $ to $ {{\sin }^{-1}}\left( \dfrac{1}{x} \right) $ .
We know that $ {{\csc }^{-1}}\left( x \right)={{\sin }^{-1}}\left( \dfrac{1}{x} \right) $ . The condition is $ x>0 $ .
For our problem value of $ x $ is 2 which is greater than 0. So, $ {{\csc }^{-1}}\left( 2 \right)={{\sin }^{-1}}\left( \dfrac{1}{2} \right) $ .
The arcus function represents the angle which on ratio sin gives the value.
So, $ \arcsin \left( x \right)={{\sin }^{-1}}x $ . If \[\arcsin \left( x \right)={{\sin }^{-1}}x=\alpha \] then we can say $ \sin \alpha =x $ .
Each of the trigonometric functions is periodic in the real part of its argument, running through all its values twice in each interval of $ 2\pi $ .
The general solution for that value where $ \sin \alpha =x $ will be $ n\pi \pm {{\left( -1 \right)}^{n}}\alpha ,n\in \mathbb{Z} $ .
But for $ \arcsin \left( x \right) $ , we won’t find the general solution. We use the principal value. For ratios sin we have $ -\dfrac{\pi }{2}\le \arcsin \left( x \right)\le \dfrac{\pi }{2} $ .
We now place the value of $ x=\dfrac{1}{2} $ in the function of $ \arcsin \left( x \right) $ .
Let the angle be $ \theta $ for which $ \arcsin \left( \dfrac{1}{2} \right)=\theta $ . This gives $ \sin \theta =\dfrac{1}{2} $ .
We know that $ \sin \theta =\dfrac{1}{2}=\sin \left( \dfrac{\pi }{6} \right) $ which gives $ \theta =\dfrac{\pi }{6} $
We get the value of y coordinates as $ \dfrac{\pi }{6} $ . Therefore, the exact value of $ {{\csc }^{-1}}\left( 2 \right) $ is $ \dfrac{\pi }{6} $ .
So, the correct answer is “ $ \dfrac{\pi }{6} $ ”.

Note: If we are finding an $ \arcsin \left( x \right) $ of a positive value, the answer is between $ 0\le \arcsin \left( x \right)\le \dfrac{\pi }{2} $ . If we are finding the $ \arcsin \left( x \right) $ of a negative value, the answer is between $ -\dfrac{\pi }{2}\le \arccos \left( x \right)\le 0 $ .