Question

How do you find the $6$ trigonometric functions for $\dfrac{{16\pi }}{3}$?

Answer 1

Hint: In order to find the trigonometric functions for the given angle, if the angle is not one of the standard values we remember from the trigonometric table, we split the angle into two standard angles whose tangent value we know, such that the sum or difference of two angles results in the given angle.

Complete step-by-step solution:
Here, we need to find the $6$ trigonometric functions, which are, $\sin $ , $\cos $ , $\tan $ , $\cot $ , $\sec $ and $\cos ec$ . Now we will find the all trigonometric functions for the given angle.
Let us take, $x = \dfrac{{16\pi }}{3}$
Now, firstly we find the value of $\sin $ function, we have
$\sin x = \sin \left( {\dfrac{{16\pi }}{3}} \right)$ , we write it as
$ = \sin \left( {\dfrac{{\pi + 15\pi }}{3}} \right)$
$ = \sin \left( {\dfrac{\pi }{3} + \dfrac{{15\pi }}{3}} \right)$ , now we simplify it and write it as
$ = \sin \left( {\dfrac{\pi }{3} + 5\pi } \right)$
$ = \sin \left( {\dfrac{\pi }{3} + \pi } \right)$ , because $\sin (2n + 1)\pi = \sin \pi $ , where $n$ is any integer, therefore we can take $5\pi $ as $\pi $
$ = \sin \left( {\pi + \dfrac{\pi }{3}} \right)$, as this angle of sine lies in third quadrant, then this expression becomes negative
$ = - \sin \dfrac{\pi }{3}$
The exact value of $\sin \dfrac{\pi }{3}$ is $\dfrac{{\sqrt 3 }}{2}$. Therefore, we get
$ = - \sin \dfrac{\pi }{3} = - \dfrac{{\sqrt 3 }}{2}$
∴ $\sin \dfrac{{16\pi }}{3} = - \dfrac{{\sqrt 3 }}{2}$ .
Similarly, we can find the other $5$ trigonometric functions. So, now we have
$\cos x = \cos \left( {\dfrac{{16\pi }}{3}} \right)$
$ = \cos \left( {\dfrac{{\pi + 15\pi }}{3}} \right)$ , we can also write it as
$ = \cos \left( {\dfrac{\pi }{3} + \dfrac{{15\pi }}{3}} \right)$
$ = \cos \left( {\dfrac{\pi }{3} + 5\pi } \right)$ , because $\cos (2n + 1)\pi = \cos \pi $ , where $n$ is any integer, therefore we can take $5\pi $ as $\pi $
$ = \cos \left( {\dfrac{\pi }{3} + \pi } \right)$ , again this angle is negative in third quadrant, then we have
$ = - \cos \dfrac{\pi }{3}$
Now, the exact value of $\cos \dfrac{\pi }{3}$ is $\dfrac{1}{2}$ . Therefore, we get
$ = - \cos \dfrac{\pi }{3} = - \dfrac{1}{2}$
∴ $\cos \dfrac{{16\pi }}{3} = - \dfrac{1}{2}$ .
Now, we need to find the value of $\tan x$ ,
We know that, $\tan x = \dfrac{{\sin x}}{{\cos x}}$
By substituting the values of $\sin x$ and $\cos x$ , we get
$ = \dfrac{{\dfrac{{ - \sqrt 3 }}{2}}}{{\dfrac{{ - 1}}{2}}}$
$ = \dfrac{{ - \sqrt 3 }}{2} \times \dfrac{{ - 2}}{1}$ , we do the reciprocal of the denominator, and after simplifying, we get
$ = \sqrt 3 $
∴ $\tan \dfrac{{16\pi }}{3} = \sqrt 3 $
Now, we know that $\cot x = \dfrac{1}{{\tan x}}$ , so we get
$ = \cot x = \dfrac{1}{{\sqrt 3 }}$
∴ $\cot \dfrac{{16\pi }}{3} = \dfrac{1}{{\sqrt 3 }}$ .
Similarly, we know that $\sec x = \dfrac{1}{{\cos x}}$ , so we get
$ = \sec x = \dfrac{1}{{\dfrac{{ - 1}}{2}}}$
$ \Rightarrow \sec x = - 2$ , so we get
∴ $\sec \dfrac{{16\pi }}{3} = - 2$ .
Now, we know that, $\cos ecx = \dfrac{1}{{\sin x}}$ , so we get
$\cos ecx = \dfrac{1}{{\dfrac{{ - \sqrt 3 }}{2}}}$
$ \Rightarrow \cos ecx = \dfrac{{ - 2}}{{\sqrt 3 }}$ so we get
∴ $\cos ec\dfrac{{16\pi }}{3} = \dfrac{{ - 2}}{{\sqrt 3 }}$ .
Hence, these are the required values of the $6$ trigonometric functions.

Note: The value of angle we evaluate here for all the trigonometric functions is in the radian measure. And, in order to find the exact value of all the trigonometric ratios we split the given angle in the two standard angles and then evaluated the values.