Question

Find the principal value of $cos^{-1}[cos(680)^o]$.

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, 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

First, we will find the value of $cos(680)^o$, and then we will further solve the cosine inverse function. We know that the period of cosine function is $360^o$ so,
$cosA^o = cos(360 + A)^o = cos(720 + A)^o$
$cos(680)^o = cos(720 + (-40))^o$
$cos(680)^o = cos(-40)^o$
We also know that $cosA = cos(-A)$ so,
$cos(-40)^o = cos40^o$

Now we will further simplify the expression as-
$cos^{-1}[cos(680)^o] = cos^{-1}(cos40^o)$

We know that $cos^{-1}(cosA) = A$, where $0 < A < 90^o$. So,
$cos^{-1}(cos40^o) = 40^o$
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. In this case, the principal value of the cosine function ranges between $0^o$ and $180^o$. This is because there can be infinite values of any inverse trigonometric functions. One common mistake by students is that they give the final answer in radian if the question has the angle in degrees or vice versa. We should always give the answer in the same units as the question.