Question

How would you find the exact value of the six trigonometric functions of $\dfrac{{5\pi }}{3}$?

Answer 1

In this question, we need to find the values of the trigonometric functions for $\dfrac{{5\pi }}{3}$. We will use the basic identities of the trigonometric functions to find out the value of the given expression. We will rewrite the terms given in the expression, as a difference of two angles in radians by applying the properties of trigonometric function. Then we make use of the difference formula of sine and cosine of the trigonometric function and simplify the equation. After that other trigonometric functions values are obtained from sine and cosine values. Then we obtain the required six trigonometric functions.

Complete step-by-step answer:
In this problem we are given the value of the angle as $\dfrac{{5\pi }}{3}$
We are asked to find the exact value of the six trigonometric functions of the above angle.
To find the values after simplification, make use of a trigonometric table or calculator, so that it reduces our work.
Firstly, we will find the value for sine.
We can write the angle $\dfrac{{5\pi }}{3}$ as follows.
$\dfrac{{5\pi }}{3} = \dfrac{{6\pi }}{3} - \dfrac{\pi }{3}$
$\Rightarrow \dfrac{{5\pi }}{3} = 2\pi - \dfrac{\pi }{3}$
Firstly, we will find the value for the sine function.
$\sin \left( {\dfrac{{5\pi }}{3}} \right) = \sin \left( {2\pi - \dfrac{\pi }{3}} \right)$
We use the formula $\sin (A - B) = \sin A\cos B - \cos A\sin B$
Here we have $A = 2\pi$ and $B = \dfrac{\pi }{3}$
Putting the values in the formula, we get,
$\Rightarrow \sin \left( {2\pi - \dfrac{\pi }{3}} \right) = \sin 2\pi \cos \dfrac{\pi }{3} - \cos 2\pi \sin \dfrac{\pi }{3}$
We know that the values of $\sin 2\pi = 0$, $\cos \dfrac{\pi }{3} = \dfrac{1}{2}$, $\cos 2\pi = 1$ and $\sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}$
Substituting this we get,
$\Rightarrow \sin \left( {2\pi - \dfrac{\pi }{3}} \right) = 0 \times \dfrac{1}{2} - 1 \times \dfrac{{\sqrt 3 }}{2}$
Simplifying this we get,
$\Rightarrow \sin \left( {2\pi - \dfrac{\pi }{3}} \right) = 0 - \dfrac{{\sqrt 3 }}{2}$
$\Rightarrow \sin \left( {2\pi - \dfrac{\pi }{3}} \right) = - \dfrac{{\sqrt 3 }}{2}$
Now we will find the value of cosine function.
$\cos \left( {\dfrac{{5\pi }}{3}} \right) = \cos \left( {2\pi - \dfrac{\pi }{3}} \right)$
We use the formula $\cos (A - B) = \cos A\cos B + \sin A\sin B$
Here we have $A = 2\pi$ and $B = \dfrac{\pi }{3}$
Putting the values in the formula, we get,
$\Rightarrow \cos \left( {2\pi - \dfrac{\pi }{3}} \right) = \cos 2\pi \cos \dfrac{\pi }{3} + \sin 2\pi \sin \dfrac{\pi }{3}$
We know that the values of $\sin 2\pi = 0$, $\cos \dfrac{\pi }{3} = \dfrac{1}{2}$, $\cos 2\pi = 1$ and $\sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}$
Substituting this we get,
$\Rightarrow \cos \left( {2\pi - \dfrac{\pi }{3}} \right) = 1 \times \dfrac{1}{2} + 0 \times \dfrac{{\sqrt 3 }}{2}$
Simplifying this we get,
$\Rightarrow \cos \left( {2\pi - \dfrac{\pi }{3}} \right) = \dfrac{1}{2} - 0$
$\Rightarrow \cos \left( {2\pi - \dfrac{\pi }{3}} \right) = \dfrac{1}{2}$
Now we will find the value of tan function.
We know that tan is a ratio of sine and cosine. Since we know the values for sine and cosine, we substitute it and obtain the tan value.
We have $\tan A = \dfrac{{\sin A}}{{\cos A}}$
Here $A = \dfrac{{5\pi }}{3}$
So we get, $\tan \dfrac{{5\pi }}{3} = \dfrac{{\sin \dfrac{{5\pi }}{3}}}{{\cos \dfrac{{5\pi }}{3}}}$
We have the values $\sin \left( {\dfrac{{5\pi }}{3}} \right) = - \dfrac{{\sqrt 3 }}{2}$ and $\cos \left( {\dfrac{{5\pi }}{3}} \right) = \dfrac{1}{2}$.
Hence substituting this we get,
$\Rightarrow \tan \left( {\dfrac{{5\pi }}{3}} \right) = \dfrac{{\dfrac{{\sqrt 3 }}{2}}}{{\dfrac{1}{2}}}$
This can be written as,
$\Rightarrow \tan \left( {\dfrac{{5\pi }}{3}} \right) = \dfrac{{\sqrt 3 }}{2} \times \dfrac{2}{1}$
$\Rightarrow \tan \left( {\dfrac{{5\pi }}{3}} \right) = \sqrt 3$.
Now we know that cot is a reciprocal of tan function. So we have,
$\Rightarrow \cot \left( {\dfrac{{5\pi }}{3}} \right) = \dfrac{1}{{\tan \left( {\dfrac{{5\pi }}{3}} \right)}}$
$\Rightarrow \cot \left( {\dfrac{{5\pi }}{3}} \right) = \dfrac{1}{{\sqrt 3 }}$
We know that secant is the reciprocal of cosine. Hence we have,
$\Rightarrow \sec \left( {\dfrac{{5\pi }}{3}} \right) = \dfrac{1}{{\cos \left( {\dfrac{{5\pi }}{3}} \right)}}$
$\Rightarrow \sec \left( {\dfrac{{5\pi }}{3}} \right) = \dfrac{1}{{\dfrac{1}{2}}}$
$\Rightarrow \sec \left( {\dfrac{{5\pi }}{3}} \right) = 2$.
Also we have cosecant is the reciprocal of sine function. So we get,
$\Rightarrow \csc \left( {\dfrac{{5\pi }}{3}} \right) = \dfrac{1}{{\sin \left( {\dfrac{{5\pi }}{3}} \right)}}$
$\Rightarrow \csc \left( {\dfrac{{5\pi }}{3}} \right) = \dfrac{1}{{\dfrac{{\sqrt 3 }}{2}}}$
$\Rightarrow \csc \left( {\dfrac{{5\pi }}{3}} \right) = \dfrac{2}{{\sqrt 3 }}$
Hence the six trigonometric functions of $\dfrac{{5\pi }}{3}$are given as, $\sin \left( {\dfrac{{5\pi }}{3}} \right) = - \dfrac{{\sqrt 3 }}{2}$, $\cos \left( {\dfrac{{5\pi }}{3}} \right) = \dfrac{1}{2}$, $\tan \left( {\dfrac{{5\pi }}{3}} \right) = \sqrt 3$, $\cot \left( {\dfrac{{5\pi }}{3}} \right) = \dfrac{1}{{\sqrt 3 }}$, $\sec \left( {\dfrac{{5\pi }}{3}} \right) = 2$ and $\csc \left( {\dfrac{{5\pi }}{3}} \right) = \dfrac{2}{{\sqrt 3 }}$.

Note:
Students must know the basic properties of trigonometric functions and also in which quadrant which function is positive or negative.
As in the first quadrant all the six trigonometric functions are positive. In the second quadrant only the sine and cosec functions are positive, rest of all are negative. In the third quadrant, only the tan and cot functions are positive and all the other functions are negative. In the fourth quadrant only the cosine and secant are positive.
The sum and difference formula related to sine and cosine are given below.
(1) $\sin (A + B) = \sin A\cos B + \cos A\sin B$
(2) $\sin (A - B) = \sin A\cos B - \cos A\sin B$
(3) $\cos (A + B) = \cos A\cos B - \sin A\sin B$
(4) $\cos (A - B) = \cos A\cos B + \sin A\sin B$