Question

Using the principals, evaluate $tan^{-1}(1) + sin^{-1}(-\dfrac{1}{2})$

Answer 1

Hint: The principal value of an inverse trigonometric function at a point x is the value of the inverse function at the point x, which lies in the range of the principal branch.We have to find the principal value of $tan^{-1}(1)$. We know that the principal value of $tan^{-1}(x)$ is given by $\left( { - \dfrac{{{\pi }}}{2},\dfrac{{{\pi }}}{2}} \right)$ . The principal value of the sine function is from $\left[ { - \dfrac{{{\pi }}}{2},\dfrac{{{\pi }}}{2}} \right]$. We will then add these values to get the answer.

Complete step-by-step answer:

The values of the tangent functions are-

Function	$0^o$	$30^o$	$45^o$	$60^o$	$90^o$
tan	0	$\dfrac{1}{{\sqrt 3 }}$	1	$\sqrt 3$	Not defined
sin	0	$\dfrac{1}{2}$	$\dfrac{1}{{\sqrt 2 }}$	$\dfrac{{\sqrt 3 }}{2}$	1

In the given question we need to find the principal value of $tan^{-1}(1) + sin^{-1}(-\dfrac{1}{2})$. The tangent and sine functions have similar range of principal values, which are positive for positive values and vice versa.

We know that $tan45^o = 1$, so
$tan^{-1}(1)= 45^o$ …(1)

Also, we know that $sin30^o = \dfrac{1}{2}$, so
$sin^{-1}(\dfrac{1}{2}) = 30^o$
So, the value of $sin^{-1}(-\dfrac{1}{2})$ can be-
$sin^{-1}(-\dfrac{1}{2}) = -30^o$ …(2)
Adding equations (1) and (2), we can get the value of the given expression as-
$tan^{-1}(1) + sin^{-1}(-\dfrac{1}{2}) = 45^o - 30^o = 15^o$
We know that ${{\pi }}$ rad = $180^o$, so
$tan^{-1}(1) + sin^{-1}(-\dfrac{1}{2}) = \dfrac{{{\pi }}}{{12}}$

This is the required value.

Note: In such types of questions, we need to strictly follow the range of the principal values that have been specified. Here, the principal values of both tangent and sine functions range from $-90^o$ to $90^o$. This is because there can be infinite values of any inverse trigonometric functions. We should keep in mind that the range principal values of sine and tangent functions appear to be the same but are slightly different. In the sine functions, the range is a closed bracket at $90^o$, whereas in tangent it is an open bracket.