
How do you find the value of $\cot (\dfrac{{2\pi }}{3})$ ?
Answer
540.3k+ views
Hint: Convert the angle into degree, and simplify cot into sine and cosine.
Complete step by step answer:
Firstly converting $\dfrac{{2\pi }}{3}$ to the degree to make it easier
$ \Rightarrow {120^ \circ }$
We know ${120^ \circ }$ is in the second quadrant. In the second quadrant, sine is positive and cosine is negative.
${120^ \circ }$ is a supplementary angle of ${60^ \circ }$.
Let’s break cot in sin and cos that is
We know cot is inverse of tan $\tan \theta = \dfrac{1}{{\cot \theta }}$ or vice-versa .
And $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
${120^ \circ }$ has the same sine value as ${60^ \circ }$ and the opposite cosine value as
$\sin ({180^ \circ } - x) = x$ and $\cos ({180^ \circ } - x) = - x$
Finding the value of sine
$ \Rightarrow \sin {120^ \circ } = \sin {60^ \circ }$
$ \Rightarrow \dfrac{{\sqrt 3 }}{2}$
Similarly for cosine
$ \Rightarrow \cos {120^ \circ } = - \cos {60^ \circ }$
$ \Rightarrow \dfrac{{ - 1}}{2}$
Putting the value of cos and sin in cot
$\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$
Put the values of cos and sine which we found above
$ \Rightarrow \dfrac{{ - 1}}{2}/\dfrac{{\sqrt 3 }}{2}$
$ \Rightarrow \dfrac{{ - 1}}{2} \cdot \dfrac{2}{{\sqrt 3 }}$
$ \Rightarrow \dfrac{{ - 1}}{{\sqrt 3 }}$
If you want to simplify this, you can rationalise the denominator .
Rationalizing the denominator, by multiplying numerator and denominator by root 3
$ \Rightarrow \dfrac{{ - 1}}{{\sqrt 3 }} \cdot \dfrac{{\sqrt 3 }}{{\sqrt 3 }}$
$ \Rightarrow - \dfrac{{\sqrt 3 }}{3}$
Thus , the value of $\cot (\dfrac{{2\pi }}{3})$ is $\dfrac{{ - 1}}{{\sqrt 3 }}$ or $ - \dfrac{{\sqrt 3 }}{3}$.
Additional information:
The answer can be in decimal form also.
The decimal form of obtained value is -0.577
Note:
You can also find the value of cosine and sine using their sum and difference identity.
The fundamental functions have reciprocal functions, which are their inverse. The reciprocal of sine is cos, the reciprocal of cosine is secant, and the reciprocal of a tangent is a cot.
Memorizing the trigonometric functions is positive or negative in a particular quadrant.
Complete step by step answer:
Firstly converting $\dfrac{{2\pi }}{3}$ to the degree to make it easier
$ \Rightarrow {120^ \circ }$
We know ${120^ \circ }$ is in the second quadrant. In the second quadrant, sine is positive and cosine is negative.
${120^ \circ }$ is a supplementary angle of ${60^ \circ }$.
Let’s break cot in sin and cos that is
We know cot is inverse of tan $\tan \theta = \dfrac{1}{{\cot \theta }}$ or vice-versa .
And $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
${120^ \circ }$ has the same sine value as ${60^ \circ }$ and the opposite cosine value as
$\sin ({180^ \circ } - x) = x$ and $\cos ({180^ \circ } - x) = - x$
Finding the value of sine
$ \Rightarrow \sin {120^ \circ } = \sin {60^ \circ }$
$ \Rightarrow \dfrac{{\sqrt 3 }}{2}$
Similarly for cosine
$ \Rightarrow \cos {120^ \circ } = - \cos {60^ \circ }$
$ \Rightarrow \dfrac{{ - 1}}{2}$
Putting the value of cos and sin in cot
$\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$
Put the values of cos and sine which we found above
$ \Rightarrow \dfrac{{ - 1}}{2}/\dfrac{{\sqrt 3 }}{2}$
$ \Rightarrow \dfrac{{ - 1}}{2} \cdot \dfrac{2}{{\sqrt 3 }}$
$ \Rightarrow \dfrac{{ - 1}}{{\sqrt 3 }}$
If you want to simplify this, you can rationalise the denominator .
Rationalizing the denominator, by multiplying numerator and denominator by root 3
$ \Rightarrow \dfrac{{ - 1}}{{\sqrt 3 }} \cdot \dfrac{{\sqrt 3 }}{{\sqrt 3 }}$
$ \Rightarrow - \dfrac{{\sqrt 3 }}{3}$
Thus , the value of $\cot (\dfrac{{2\pi }}{3})$ is $\dfrac{{ - 1}}{{\sqrt 3 }}$ or $ - \dfrac{{\sqrt 3 }}{3}$.
Additional information:
The answer can be in decimal form also.
The decimal form of obtained value is -0.577
Note:
You can also find the value of cosine and sine using their sum and difference identity.
The fundamental functions have reciprocal functions, which are their inverse. The reciprocal of sine is cos, the reciprocal of cosine is secant, and the reciprocal of a tangent is a cot.
Memorizing the trigonometric functions is positive or negative in a particular quadrant.
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