Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

How do you create a 16 point unit circle that ranges from 0 to 8pi?

seo-qna
SearchIcon
Answer
VerifiedVerified
422.4k+ views
Hint: Create the circle of range $\left( 0,2\pi \right)$ instead of $\left( 0,8\pi \right)$ as both are same in case of a circle and $2\pi $is the minimum range for a circle. Plot 16 points by choosing 16 different angles. Choose the angles in such a way that their sine and cosine values are known.

Complete step by step answer:
A circle that ranges from 0 to $8\pi $, is the same as the circle ranges from 0 to $2\pi $because each rotation from one quadrant to another around the coordinate axes is $\dfrac{\pi }{2}$. For a complete circle it will be a complete rotation around 4 quadrants. Hence range will be $\dfrac{\pi }{2}\times 4=2\pi $.
So, to create a circle that ranges from 0 to $8\pi $, we have to create a circle ranging from 0 to $2\pi $.
Now for 16 points we have to consider 16 different angles.
The sine and cosine values of the angles can be taken from the given table
angleCosine valueSine value
${{0}^{\circ }}$ 01
${{30}^{\circ }}$ $\dfrac{\sqrt{3}}{2}$ $\dfrac{1}{2}$
${{45}^{\circ }}$ $\dfrac{1}{\sqrt{2}}$ $\dfrac{1}{\sqrt{2}}$
${{60}^{\circ }}$$\dfrac{1}{2}$$\dfrac{\sqrt{3}}{2}$
${{90}^{\circ }}$ 10

For other angles we just have to change the sign as per the quadrant.
So, the points are
$\begin{align}
  & \left( \cos {{0}^{\circ }},\sin {{0}^{\circ }} \right)=\left( 1,0 \right) \\
 & \left( \cos {{30}^{\circ }},\sin {{30}^{\circ }} \right)=\left( \dfrac{\sqrt{3}}{2},\dfrac{1}{2} \right) \\
 & \left( \cos {{45}^{\circ }},\sin {{45}^{\circ }} \right)=\left( \dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}} \right) \\
 & \left( \cos {{60}^{\circ }},\sin {{60}^{\circ }} \right)=\left( \dfrac{1}{2},\dfrac{\sqrt{3}}{2} \right) \\
 & \left( \cos {{90}^{\circ }},\sin {{90}^{\circ }} \right)=\left( 0,1 \right) \\
 & \left( \cos {{120}^{\circ }},\sin {{120}^{\circ }} \right)=\left( -\dfrac{1}{2},\dfrac{\sqrt{3}}{2} \right) \\
 & \left( \cos {{135}^{\circ }},\sin {{135}^{\circ }} \right)=\left( -\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}} \right) \\
 & \left( \cos {{150}^{\circ }},\sin {{150}^{\circ }} \right)=\left( -\dfrac{\sqrt{3}}{2},\dfrac{1}{2} \right) \\
 & \left( \cos {{180}^{\circ }},\sin {{180}^{\circ }} \right)=\left( -1,0 \right) \\
 & \left( \cos {{210}^{\circ }},\sin {{210}^{\circ }} \right)=\left( -\dfrac{\sqrt{3}}{2},-\dfrac{1}{2} \right) \\
 & \left( \cos {{225}^{\circ }},\sin {{225}^{\circ }} \right)=\left( -\dfrac{1}{\sqrt{2}},-\dfrac{1}{\sqrt{2}} \right) \\
 & \left( \cos {{240}^{\circ }},\sin {{240}^{\circ }} \right)=\left( -\dfrac{1}{2},-\dfrac{\sqrt{3}}{2} \right) \\
 & \left( \cos {{270}^{\circ }},\sin {{270}^{\circ }} \right)=\left( 0,-1 \right) \\
 & \left( \cos {{300}^{\circ }},\sin {{300}^{\circ }} \right)=\left( \dfrac{1}{2},-\dfrac{\sqrt{3}}{2} \right) \\
 & \left( \cos {{315}^{\circ }},\sin {{315}^{\circ }} \right)=\left( \dfrac{1}{\sqrt{2}},-\dfrac{1}{\sqrt{2}} \right) \\
 & \left( \cos {{330}^{\circ }},\sin {{330}^{\circ }} \right)=\left( \dfrac{\sqrt{3}}{2},-\dfrac{1}{2} \right) \\
 & \left( \cos {{360}^{\circ }},\sin {{360}^{\circ }} \right)=\left( 1,0 \right) \\
\end{align}$
seo images

Note:
The angels should be chosen in such a way that their sine and cosine values are known or can be obtained easily. The 16 equal angle difference i.e. $\dfrac{{{360}^{\circ }}}{16}={{22.5}^{\circ }}$should be avoided because there will be complexity in calculation. For sine and cosine values of angles more than ${{90}^{\circ }}$, ASTC rule can be used.