How would you find the exact trigonometric ratio for an angle whose radian measure is \[\dfrac{4\pi }{3}\]?

Hint: To find the trigonometric ratios, we will use the following properties. \[\sin (\pi +x)=-\sin x\], \[\cos (\pi +x)=-\cos x\]and \[\tan (\pi +x)=\tan x\]where x is an angle measured in radian. We should also know that \[\csc x=\dfrac{1}{\sin x}\], \[\sec x=\dfrac{1}{\cos x}\]and \[\cot x=\dfrac{1}{\tan x}\].

Complete step by step answer:
We are given an angle of radian measure \[\dfrac{4\pi }{3}\], and we have to find the value of the trigonometric ratios, that is the value of\[\sin \left( \dfrac{4\pi }{3} \right),\cos \left( \dfrac{4\pi }{3} \right),\tan \left( \dfrac{4\pi }{3} \right),\csc \left( \dfrac{4\pi }{3} \right),\sec \left( \dfrac{4\pi }{3} \right)\And \cot \left( \dfrac{4\pi }{3} \right)\]. As the measure of the angle is \[\dfrac{4\pi }{3}\], it is a third quadrant angle. The angles in the third quadrant are of form \[\pi +x\] where\[0\le x\le \dfrac{\pi }{2}\]we can find the value of x by comparing this with \[\dfrac{4\pi }{3}\]
\[\Rightarrow \pi +x=\dfrac{4\pi }{3}\]
Subtracting \[\pi \]from both sides of the above equation, we get
\[\Rightarrow \pi +x-\pi =\dfrac{4\pi }{3}-\pi \]
\[\Rightarrow x=\dfrac{\pi }{3}\]
We also know the trigonometric identities which states that \[\sin (\pi +x)=-\sin x\], \[\cos (\pi +x)=-\cos x\]and \[\tan (\pi +x)=\tan x\]. Here we have \[x=\dfrac{\pi }{3}\].
Using the above relations and values, we can find the value of \[\sin \left( \dfrac{4\pi }{3} \right),\cos \left( \dfrac{4\pi }{3} \right)\]and \[\tan \left( \dfrac{4\pi }{3} \right)\] as follows,
\[\begin{align}
  & \Rightarrow \sin \left( \dfrac{4\pi }{3} \right)=\sin \left( \pi +\dfrac{\pi }{3} \right)=-\sin \left( \dfrac{\pi }{3} \right)=-\dfrac{\sqrt{3}}{2} \\
 & \Rightarrow \cos \left( \dfrac{4\pi }{3} \right)=\cos \left( \pi +\dfrac{\pi }{3} \right)=-\cos \left( \dfrac{\pi }{3} \right)=-\dfrac{1}{2} \\
 & \Rightarrow \tan \left( \dfrac{4\pi }{3} \right)=\sin \left( \pi +\dfrac{\pi }{3} \right)=-\sin \left( \dfrac{\pi }{3} \right)=\sqrt{3} \\
\end{align}\]
We also know that the relation between trigonometric ratios which states that, \[\csc x=\dfrac{1}{\sin x},\sec x=\dfrac{1}{\cos x}\]and \[\cot x=\dfrac{1}{\tan x}\]. Using these relations and the values, we can find the other trigonometric ratios as follows,
\[\begin{align}
  & \Rightarrow \csc \left( \dfrac{4\pi }{3} \right)=\dfrac{1}{\sin \left( \dfrac{4\pi }{3} \right)}=\dfrac{1}{-\dfrac{\sqrt{3}}{2}}=\dfrac{-2}{\sqrt{3}} \\
 & \Rightarrow \sec \left( \dfrac{4\pi }{3} \right)=\dfrac{1}{\cos \left( \dfrac{4\pi }{3} \right)}=\dfrac{1}{-\dfrac{1}{2}}=-2 \\
 & \Rightarrow \cot \left( \dfrac{4\pi }{3} \right)=\dfrac{1}{\tan \left( \dfrac{4\pi }{3} \right)}=\dfrac{1}{\sqrt{3}} \\
\end{align}\]
Hence the values of the trigonometric ratios are \[\sin \left( \dfrac{4\pi }{3} \right)=-\dfrac{\sqrt{3}}{2}\], \[\cos \left( \dfrac{4\pi }{3} \right)=-\dfrac{1}{2}\], \[\tan \left( \dfrac{4\pi }{3} \right)=\sqrt{3}\], \[\csc \left( \dfrac{4\pi }{3} \right)=\dfrac{-2}{\sqrt{3}}\], \[\sec \left( \dfrac{4\pi }{3} \right)=-2\]and \[\cot \left( \dfrac{4\pi }{3} \right)=\dfrac{1}{\sqrt{3}}\]for the angle of radian measure \[\dfrac{4\pi }{3}\].

Note:
These types of problems are can be easily solved by remembering basic trigonometric identifies like \[T\left( \dfrac{\pi }{2}\pm x \right)\], \[T\left( \pi \pm x \right)\], \[T\left( \dfrac{3\pi }{2}\pm x \right)\] and \[T\left( 2\pi \pm x \right)\]where T represents one if the trigonometric ratio. One should also remember that, which trigonometric ratios are positive in a particular quadrant.