Question

How do you find the exact value of $\sec \left( { - \dfrac{\pi }{3}} \right)$?

Answer 1

Hint:The negative sign given in question ; that means it is measured in a clockwise direction. We shall try to get the equivalent positive angle (measured in a counter-clockwise direction) to solve the problem.
Also, we have to keep in mind that the secant function is positive in the first and fourth quadrant.

Formula used: $1radian = \dfrac{{180^\circ }}{\pi }$

Complete step-by-step answer:
First of all, let us find the equivalent positive angle of $\left( {\dfrac{{ - \pi }}{3}} \right)$ .
Since the magnitude of the angle is less than $2\pi $, we can find the required angle by adding $2\pi $.
That is,
$\left( {\dfrac{{ - \pi }}{3}} \right) \Leftrightarrow \left( {2\pi + \dfrac{{ - \pi }}{3}} \right) = \left( {\dfrac{{5\pi }}{3}} \right)$
That is the equivalent positive angle (measure anti-clockwise) is $\left( {\dfrac{{5\pi }}{3}} \right)radians$ .
Now, let us convert this into a degree measure using the formula $1radian = \dfrac{{180^\circ }}{\pi }$ .
$ \Rightarrow \left( {\dfrac{{5\pi }}{3}} \right)radians = \dfrac{{5\pi }}{3} \times \dfrac{{180^\circ }}{\pi } = 300^\circ $
Now, $300^\circ $ lies in the ${4^{th}}$quadrant and hence its secant value is positive.
Therefore, we have
$\sec \left( { - \dfrac{\pi }{3}} \right) = \sec \left( {300^\circ } \right)$
$ \Rightarrow \sec \left( {300^\circ } \right) = \sec \left( {360^\circ - 60^\circ } \right)$
Now, we know that the period of secant function is $2\pi $ or $360^\circ $.
Therefore,
$\sec \left( {360^\circ - 60^\circ } \right) = \sec \left( {360^\circ + \left( { - 60^\circ } \right)} \right) = \sec \left( { - 60^\circ } \right)$
Now, we know that $\sec \theta = \dfrac{1}{{\cos \theta }}$
Also, we know that $\cos \left( { - \theta } \right) = \cos \left( \theta \right)$
$ \Rightarrow \sec \left( { - \theta } \right) = \dfrac{1}{{\cos \left( { - \theta } \right)}} = \dfrac{1}{{\cos \left( \theta \right)}} = \sec \left( \theta \right)$
$ \Rightarrow \sec \left( { - 60^\circ } \right) = \sec \left( {60^\circ } \right)$
As we all know, $\sec \left( {60^\circ } \right) = 2$
Therefore,
$\sec \left( {\dfrac{{ - \pi }}{3}} \right) = 2$ , which is our final answer.

Additional information:
If the angle within the trigonometric function is greater than $360^\circ $, then we have to remove full rotations of $360^\circ $ until finally, we obtain an angle $\theta $, such that
$0 \leqslant \theta \leqslant 2\pi $.

Note: The negative angle inside a trigonometric function can be removed easily using the following properties,
$\sin \left( { - x} \right) = - \sin x$
$\cos \left( { - x} \right) = \cos x$
For the rest of the ratios, these two relations can be used.