Question

What is $\sin \left( \dfrac{\pi }{3} \right)$ of unit circle?

Answer 1

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

Complete step by step solution:
We need to find the ratio value for $\sin \left( \dfrac{\pi }{3} \right)$ .
For general form of $\sin \left( x \right)$ , 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 sin and if it’s odd then the ratio changes to cos ratio from sin.
Then we find the position of the given angle as quadrant value measured in counter clockwise movement from the origin and the positive side of X-axis.
If the angel 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.
Depending on the sign and ratio change the final angle becomes $\alpha$ from x.
The final form becomes $\sin \left( \dfrac{\pi }{3} \right)=\sin \left( 0\times \dfrac{\pi }{2}+\dfrac{\pi }{3} \right)=\sin \left( \dfrac{\pi }{3} \right)=\dfrac{\sqrt{3}}{2}$ .
Therefore, the value of $\sin \left( \dfrac{\pi }{3} \right)$ is $\dfrac{\sqrt{3}}{2}$ .
So, the correct answer is “ $\dfrac{\sqrt{3}}{2}$ ”.

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.