Question

How do you find the value of $\csc \left( -240 \right)$?

Answer 1

We explain the process of finding values for associated angles. We find the rotation and the position of the angle for $\left( -240 \right)$. We explain the changes that are required for that angle. Depending on those things we find the solution.

Complete answer:
We need to find the ratio value for $\csc \left( -240 \right)$.
For general form of $\csc \left( x \right)$, 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 $\left| \alpha \right|\le \dfrac{\pi }{2}$.
Now we take the value of k. If it’s even then keep the ratio as cosec and if it’s odd then the ratio changes to sec ratio from cosec.
Then we find the position of the given angle as a quadrant value measured in counter clockwise movement from the origin and the positive side of the X-axis.
If the angle falls in the first or second quadrant then the sign remains positive but if it falls in the third or fourth quadrant then the sign becomes negative.
For the given angle $-{{240}^{\circ }}$, we can express as $240=-\left( 2\times \dfrac{\pi }{2}+60 \right)$. The negative sign can be omitted at the end as it is just to show the direction and position of the angle with respect to the quadrants.
The value of k is even which means the trigonometrical ratio doesn’t change.
The position of the angle is in the second quadrant. The angle completes the half-circle 1 times and then goes $\dfrac{\pi }{3}$ clockwise. Therefore, the sign remains positive.
The final form becomes $\csc \left( -240 \right)=\csc \left( 2\times \dfrac{\pi }{2}+60 \right)=\csc \left( 60 \right)=\dfrac{2}{\sqrt{3}}$.
Therefore, the value of $\csc \left( -240 \right)$ is $\dfrac{2}{\sqrt{3}}$.

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.