Question

Find the value of ${\sin ^{ - 1}}(\sin 5) =$
A. $5$
B. $5 - 2\pi$
C. $2\pi - 5$
D. $2\pi + 5$

Answer 1

Hint: We know that ${\sin ^{ - 1}}(\sin \theta ) = \theta$ provided that. Since we have written the domain and range check where $5$ lies in that range. The given value is clearly above the range so find where $5$ exists, i.e. in which quadrant, and then evaluate. Check the periodicity of the trigonometric function to make the given value fit in the range.

Complete step by step solution:
The inverse function of $\sin \theta$ is ${\sin ^{ - 1}}x$
We can deduce a formula including both the functions to get,
$\Rightarrow {\sin ^{ - 1}}(\sin \theta ) = \theta$
The range and domain are given by or
Since in the question it is given as $\theta = 5$ we can see that it is outside the range.
It lies in the range of $\left( {\dfrac{{3\pi }}{2},2\pi } \right)$ which is the fourth quadrant.
In this quadrant $\sin$ is negative.
We also know that $\sin$ is a periodic function. So, we can neglect multiples by subtracting $2\pi$ .
Here $2\pi$ is the general period of the $\sin$ function.
So, we subtract it with $2\pi$ (since $5 - 2\pi > - \dfrac{\pi }{2}$ )
$\Rightarrow {\sin ^{ - 1}}(\sin 5) = 5 - 2\pi$

$\therefore$ The solution for ${\sin ^{ - 1}}(\sin 5)$ is $5 - 2\pi$ which is option B

Additional Information: The inverse functions in trigonometry are also known as arc functions or anti trigonometric functions. They are majorly known as arc functions because they are most used to find the length of the arc needed to get the given or specified value. We can convert a function into an inverse function and vice versa.

Note: Check where the trigonometric functions become negative or positive. Also, whenever the value is out of range or domain check the function’s periodicity and then subtract or add it with the general period to get it back into the range. Always check when the trigonometric functions are given in degrees or radians.