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# Find the value of $\cos \left( \dfrac{2\pi }{3} \right)$?

Last updated date: 21st Jul 2024
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Hint: We explain the process of finding values for associated angles. We find the rotation and the position of the angle for $\left( \dfrac{2\pi }{3} \right)$. We explain the changes that are required for that angle. Depending on those things we find the solution.

We need to find the ratio value for $\cos \left( \dfrac{2\pi }{3} \right)$.
For general form of $\cos \left( x \right)$, we need to convert the value of x into the closest multiple of $\dfrac{\pi }{2}$ and add or subtract a certain value $\alpha$ from that multiple of $\dfrac{\pi }{2}$ to make it equal to x.
Let’s assume $x=k\times \dfrac{\pi }{2}+\alpha$, $k\in \mathbb{Z}$. Here we took the addition of $\alpha$. We also need to remember that $\left| \alpha \right|\le \dfrac{\pi }{2}$.
The final form becomes $\cos \left( \dfrac{2\pi }{3} \right)=\cos \left( 1\times \dfrac{\pi }{2}+\dfrac{\pi }{6} \right)=-\sin \left( \dfrac{\pi }{6} \right)=-\dfrac{1}{2}$.
Therefore, the value of $\cos \left( \dfrac{2\pi }{3} \right)$ is $-\dfrac{1}{2}$.
Note: We need to remember that the easiest way to avoid the change of ratio thing is to form the multiple of $\pi$ instead of $\dfrac{\pi }{2}$. It makes the multiplied number always even. In that case we don’t have to change the ratio. If $x=k\times \pi +\alpha =2k\times \dfrac{\pi }{2}+\alpha$. Value of $2k$ is always even.