Question

How do you solve and find the value of ${{\sin }^{-1}}\left( \dfrac{1}{2} \right)$?

Answer 1

Hint: To solve the given inverse trigonometric expression i.e. ${{\sin }^{-1}}\left( \dfrac{1}{2} \right)$ we are going to assume this expression as $\theta $ and then take the sine on both the sides of this equation. After that, we will require this property that $\sin \left( {{\sin }^{-1}} \right)$ is 1. And also, we need the information that $\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}$.

Complete step by step answer:
The inverse trigonometric expression that we have to solve and find the value of:
${{\sin }^{-1}}\left( \dfrac{1}{2} \right)$
Now, let us assume that the above inverse trigonometric expression is equal to $\theta $ so writing this statement in the mathematical form we get,
${{\sin }^{-1}}\left( \dfrac{1}{2} \right)=\theta $
Taking sine on both the sides we get,
$\sin \left( {{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right)=\sin \theta $
We know that there is an algebraic property that if a number or expression is multiplied by its inverse then we will get 1 so $\sin \left( {{\sin }^{-1}} \right)$ is equal to 1 in the above expression.
$\dfrac{1}{2}=\sin \theta $
We know that from the angles of sine that:
$\sin \dfrac{\pi }{6}=\dfrac{1}{2}$
On comparing this trigonometric value with $\dfrac{1}{2}=\sin \theta $ we get,
$\theta =\dfrac{\pi }{6}$
And the general solution for this angle is equal to:
$\theta =2n\pi \pm \dfrac{\pi }{6}$
The value that “n” takes in the above equation is from 1, 2, 3……..

Note:
A concept that you can remember here is that the domain of ${{\sin }^{-1}}x$ contains value from -1 to 1 so if you see any value of x which is outside this domain then the solution won’t exist for ${{\sin }^{-1}}x$. Sometimes, examiner gives the value of x as $\dfrac{\pi }{2},\pi $ then as the values of these x values are 1.57 and 3.14 and these values are lying outside the domain of x which is from -1 to 1 so such values of ${{\sin }^{-1}}x$ does not exist.
${{\sin }^{-1}}\left( \dfrac{\pi }{2} \right),{{\sin }^{-1}}\pi $
The above inverse trigonometric expressions won’t exist.