Question

How do you evaluate \[\sec (\dfrac{\pi }{3})\].

Answer 1

Hint:
This sum can be solved directly as it is a common value of the $\cos $function. In order to solve such a type of sum with values not common for example $\cos (120)$a student has to make use of the formula of $\cos $to evaluate. To evaluate the above sum the student can substitute the value directly and inverse the answer as $\sec $is inverse of$\cos $ or can use the formula $\cos (A - B) = \cos A\cos B + \sin A\sin B$as we know the value of $\cos (90),\cos (30),\sin (90),\sin (30)$. Another way to solve such a type of sum is by proving it.

Complete step by step solution:
The 1st Method is directly using the value and noting down the answer. Since $\pi $stands for ${180^ \circ }$and it is divided by $3$,$\sec (\dfrac{\pi }{3})$ is equivalent to $\dfrac{1}{{\cos ({{60}^ \circ })}}$.
$\therefore \cos ({60^ \circ }) = \dfrac{1}{2}$
\[\therefore \sec ({60^ \circ }) = \dfrac{1}{{\cos ({{60}^ \circ })}} = 2\]
2nd Method is to use the formula $\cos (A - B) = \cos A\cos B + \sin A\sin B$ and finding the answer for $\cos ({60^ \circ })$and finding the inverse of this value in order to find $\sec (\dfrac{\pi }{3})$
We can write $\cos ({60^ \circ })$ as $\cos ({90^ \circ } - {30^ \circ })$.
Using the formula :
$\cos ({90^ \circ } - {30^ \circ }) = \cos 90 \times \cos 30 + \sin 30 \times \sin 90..........(1)$
As we know that value of $\cos 90$is $0$,
$\cos ({90^ \circ } - {30^ \circ }) = \sin 90 \times \sin 30..........(2)$
Making use of the standard values , we get the value of $\sin ({30^ \circ })$
$\cos ({90^ \circ } - {30^ \circ }) = 1 \times \dfrac{1}{2}.........(3)$
$\therefore \cos ({30^ \circ }) = \dfrac{1}{2}$
\[\therefore \sec ({60^ \circ }) = \dfrac{1}{{\cos ({{60}^ \circ })}} = 2\]

Note:
In order to solve trigonometric numerical the student should know the standard values of all the trigonometric functions which are ${0^ \circ },{30^ \circ },{45^ \circ },{60^ \circ },{90^ \circ }$. These values would not only be useful for trigonometric related problems but also help in calculating the values of sides in triangles, useful in the heights and distances chapter. Also, they are useful while solving chapters like integration, derivatives. Thus students are advised to memorize the standard values of trigonometric functions and its inverse as well.