Question

Find the principal value of ${\cos ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 2 }}} \right)$.

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. The principal value of the cosine function is from $\left[ {0,{{\pi }}} \right]$. From the table we will find the value of the function within the principal range. Then we will use the formula
$cos^{-1}(-x) = 180^o - cos^{-1}(x)$ to get our final answer.

Complete step-by-step answer:
The values of the cosine function are-

Function	$0^o$	$30^o$	$45^o$	$60^o$	$90^o$
cos	1	$\dfrac{{\sqrt 3 }}{2}$	$\dfrac{1}{{\sqrt 2 }}$	$\dfrac{1}{2}$	0

In the given question we need to find the principal value of ${\cos ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 2 }}} \right)$. We know that for cosine function to be negative, the angle should be obtuse, that is greater than $90^o$.

We know that $\cos {45^o} = \dfrac{1}{{\sqrt 2 }}$. This means that the value of ${\cos ^{ - 1}}\left( {\dfrac{1}{{\sqrt 2 }}} \right) = {45^o}$. So the value of the given expression can be calculated as-

${\cos ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 2 }}} \right) = {180^{\text{o}}} - {45^{\text{o}}} = {135^{\text{o}}}$
We know that ${{\pi }}$ rad = $180^o$, so
${\cos ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 2 }}} \right) = \dfrac{{3{{\pi }}}}{4}$
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. The principal value of cosine in the question should always be between $0^o$ and $180^o$. This is because there can be infinite values of any inverse trigonometric functions. One common mistake is that students get confused between the sine and the cosine functions, and often write the values of sine function or vice versa. Also, students often do not pay attention to the negative, which leads to the wrong answer.