Question

How do you find all six trigonometric functions of 240 degrees?

Answer 1

Hint : Here in the question to find the value of all six trigonometric functions of ${240^ \circ }$ by using the addition and subtraction with different standard angles. Then by applying ASTC rule of trigonometry reduce the given angle ${240^ \circ }$ to the standard angle then by that value of standard angle of trigonometric function we get the required value.

Complete step-by-step answer :
ASTC rule stands for the "all sine tangent cosine" rule. It is intended to remind us that all trigonometric ratios are positive in the first quadrant of a graph, only the sine and its cofunction cosecant are positive in the second quadrant, only the tangent and its cofunction cotangent are positive in the third quadrant, and only the cosine and its cofunction secant are positive in the fourth quadrant. One way to remember this arrangement is with a sentence “All students take coffee” or “All science teachers are crazy”.
Then always remember, when you write the trigonometric function with angle ${90^ \circ }$ or ${270^ \circ }$ , the function will change to its cofunction.
The value of the all six trigonometric functions of 240 degrees are:

i.Sine of ${240^ \circ }$
$\sin \left( {{{240}^ \circ }} \right)$ can be written as $\sin \left( {{{240}^ \circ }} \right) = \sin {\left( {180 + 60} \right)^ \circ }$
The angle ${240^ \circ }$ is greater than ${180^ \circ }$ and thus lies in the third quadrant. sine functions are not positive here, hence they are negative. Here we must keep the function as sin itself.
$\Rightarrow \sin \left( {{{240}^ \circ }} \right) = \sin {\left( {180 + 60} \right)^ \circ }$
$\Rightarrow \sin \left( {{{240}^ \circ }} \right) = - \sin \left( {{{60}^ \circ }} \right)$
The value of $\sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2}$
$\therefore \sin \left( {{{240}^ \circ }} \right) = - \dfrac{{\sqrt 3 }}{2}$

ii.Cosine of ${240^ \circ }$
$\cos \left( {{{240}^ \circ }} \right)$ can be written as $\cos \left( {{{240}^ \circ }} \right) = \cos {\left( {180 + 60} \right)^ \circ }$
The angle ${240^ \circ }$ is greater than ${180^ \circ }$ and thus lies in the third quadrant. cosine functions are not positive here, hence they are negative. Here we must keep the function as cosine itself.
$\Rightarrow \cos \left( {{{240}^ \circ }} \right) = \cos {\left( {180 + 60} \right)^ \circ }$
$\Rightarrow \cos \left( {{{240}^ \circ }} \right) = - \cos \left( {{{60}^ \circ }} \right)$
The value of $\cos {60^ \circ } = \dfrac{1}{2}$
$\therefore \cos \left( {{{240}^ \circ }} \right) = - \dfrac{1}{2}$

iii.Tangent of ${240^ \circ }$
By the definition of tan x can be written as $\tan x = \dfrac{{\sin x}}{{\cos x}}$
$\Rightarrow \tan {240^0} = \dfrac{{\sin {{240}^0}}}{{\cos {{240}^0}}}$
$\Rightarrow \tan {240^0} = \dfrac{{ - \dfrac{{\sqrt 3 }}{2}}}{{ - \dfrac{1}{2}}}$
$\Rightarrow \tan {240^0} = - \dfrac{{\sqrt 3 }}{2} \times - \dfrac{2}{1}$
$\therefore \tan \left( {{{240}^0}} \right) = \sqrt 3$

iv.Cosecant of ${240^ \circ }$
By the definition of csc x can be written as $\csc x = \dfrac{1}{{\sin x}}$
$\Rightarrow \csc {240^0} = \dfrac{1}{{\sin {{240}^0}}}$
$\Rightarrow \csc {240^0} = \dfrac{1}{{ - \dfrac{{\sqrt 3 }}{2}}}$
$\therefore \csc \left( {{{240}^0}} \right) = - \dfrac{2}{{\sqrt 3 }}$

v.Secant of ${240^ \circ }$
By the definition of sec x can be written as $\sec x = \dfrac{1}{{\cos x}}$
$\Rightarrow \sec {240^0} = \dfrac{1}{{\cos {{240}^0}}}$
$\Rightarrow \sec {240^0} = \dfrac{1}{{ - \dfrac{1}{2}}}$
$\therefore \sec \left( {{{240}^0}} \right) = - 2$

v.Cotangent of ${240^ \circ }$
By the definition of cot x can be written as $\cot x = \dfrac{1}{{\tan x}}$
$\Rightarrow \cot {240^0} = \dfrac{1}{{\tan {{240}^0}}}$
$\therefore \cot \left( {{{240}^0}} \right) = \dfrac{1}{{\sqrt 3 }}$
Hence, the value of all six trigonometric functions at ${240^ \circ }$ are
$\sin \left( {{{240}^ \circ }} \right) = - \dfrac{{\sqrt 3 }}{2}$ , $\cos \left( {{{240}^ \circ }} \right) = - \dfrac{1}{2}$ , $\tan \left( {{{240}^0}} \right) = \sqrt 3$ , $\csc \left( {{{240}^0}} \right) = - \dfrac{2}{{\sqrt 3 }}$ , $\sec \left( {{{240}^0}} \right) = - 2$ and $\cot \left( {{{240}^0}} \right) = \dfrac{1}{{\sqrt 3 }}$ .

Note : The ASTC rule defined as all sine tan cosine this explains all trigonometry ratios are positive in the first quadrant. Sine trigonometry ratio is positive in the second quadrant. Tan trigonometry ratio is positive in the third quadrant. The cosine trigonometry ratio is positive in the fourth quadrant.to determine the values of angles we need to know about the table of trigonometry ratios for standard angles.