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

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

Explanation

Solution

Hint : We convert the inverse function from csc⁡−1(x){{\csc }^{-1}}\left( x \right) to sin⁡−1(1x){{\sin }^{-1}}\left( \dfrac{1}{x} \right) . Then we explain the function arcsin⁡(x)\arcsin \left( x \right) . We express the inverse function of sin in the form of arcsin⁡(x)=sin⁡−1x\arcsin \left( x \right)={{\sin }^{-1}}x . We draw the graph of arcsin⁡(x)\arcsin \left( x \right) and the line x=12x=\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(x){{\csc }^{-1}}\left( x \right) to sin⁡−1(1x){{\sin }^{-1}}\left( \dfrac{1}{x} \right) .
We know that csc⁡−1(x)=sin⁡−1(1x){{\csc }^{-1}}\left( x \right)={{\sin }^{-1}}\left( \dfrac{1}{x} \right) . The condition is x>0x>0 .
For our problem value of xx is 2 which is greater than 0. So, csc⁡−1(2)=sin⁡−1(12){{\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⁡(x)=sin⁡−1x\arcsin \left( x \right)={{\sin }^{-1}}x . If arcsin⁡(x)=sin⁡−1x=α\arcsin \left( x \right)={{\sin }^{-1}}x=\alpha then we can say sin⁡α=x\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 2π2\pi .
The general solution for that value where sin⁡α=x\sin \alpha =x will be nπ±(−1)nα,n∈Zn\pi \pm {{\left( -1 \right)}^{n}}\alpha ,n\in \mathbb{Z} .
But for arcsin⁡(x)\arcsin \left( x \right) , we won’t find the general solution. We use the principal value. For ratios sin we have −π2≤arcsin⁡(x)≤π2-\dfrac{\pi }{2}\le \arcsin \left( x \right)\le \dfrac{\pi }{2} .
We now place the value of x=12x=\dfrac{1}{2} in the function of arcsin⁡(x)\arcsin \left( x \right) .
Let the angle be θ\theta for which arcsin⁡(12)=θ\arcsin \left( \dfrac{1}{2} \right)=\theta . This gives sin⁡θ=12\sin \theta =\dfrac{1}{2} .
We know that sin⁡θ=12=sin⁡(π6)\sin \theta =\dfrac{1}{2}=\sin \left( \dfrac{\pi }{6} \right) which gives θ=π6\theta =\dfrac{\pi }{6}
We get the value of y coordinates as π6\dfrac{\pi }{6} . Therefore, the exact value of csc⁡−1(2){{\csc }^{-1}}\left( 2 \right) is π6\dfrac{\pi }{6} .
So, the correct answer is “ π6\dfrac{\pi }{6} ”.

Note : If we are finding an arcsin⁡(x)\arcsin \left( x \right) of a positive value, the answer is between 0≤arcsin⁡(x)≤π20\le \arcsin \left( x \right)\le \dfrac{\pi }{2} . If we are finding the arcsin⁡(x)\arcsin \left( x \right) of a negative value, the answer is between −π2≤arccos⁡(x)≤0-\dfrac{\pi }{2}\le \arccos \left( x \right)\le 0 .