Question

identify the points on the unit circle shown above.
a:()
b:()
c:()
d:()
e:()
f:()
g:()
h:()
i:()
j:()
k:()
l:()
m:()
n:()
o:()
p:()

Explanation:

Step1: Recall unit - circle coordinates

On the unit circle \(x = \cos\theta\) and \(y=\sin\theta\).

• For point \(A\): \(\theta = 0\), \(\cos(0)=1\), \(\sin(0) = 0\).
• For point \(B\): \(\theta=\frac{\pi}{12}\), \(\cos(\frac{\pi}{12})=\frac{\sqrt{6}+\sqrt{2}}{4}\approx0.966\), \(\sin(\frac{\pi}{12})=\frac{\sqrt{6}-\sqrt{2}}{4}\approx0.259\).
• For point \(C\): \(\theta=\frac{\pi}{6}\), \(\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}\), \(\sin(\frac{\pi}{6})=\frac{1}{2}\).
• For point \(D\): \(\theta=\frac{\pi}{4}\), \(\cos(\frac{\pi}{4})=\frac{\sqrt{2}}{2}\), \(\sin(\frac{\pi}{4})=\frac{\sqrt{2}}{2}\).
• For point \(E\): \(\theta=\frac{\pi}{2}\), \(\cos(\frac{\pi}{2}) = 0\), \(\sin(\frac{\pi}{2})=1\).
• For point \(F\): \(\theta=\frac{5\pi}{12}\), \(\cos(\frac{5\pi}{12})=\frac{\sqrt{6}-\sqrt{2}}{4}\approx0.259\), \(\sin(\frac{5\pi}{12})=\frac{\sqrt{6}+\sqrt{2}}{4}\approx0.966\).
• For point \(G\): \(\theta=\frac{\pi}{3}\), \(\cos(\frac{\pi}{3})=\frac{1}{2}\), \(\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}\).
• For point \(H\): \(\theta=\frac{2\pi}{3}\), \(\cos(\frac{2\pi}{3})=-\frac{1}{2}\), \(\sin(\frac{2\pi}{3})=\frac{\sqrt{3}}{2}\).
• For point \(I\): \(\theta=\pi\), \(\cos(\pi)=- 1\), \(\sin(\pi)=0\).
• For point \(J\): \(\theta=\frac{4\pi}{3}\), \(\cos(\frac{4\pi}{3})=-\frac{1}{2}\), \(\sin(\frac{4\pi}{3})=-\frac{\sqrt{3}}{2}\).
• For point \(K\): \(\theta=\frac{5\pi}{6}\), \(\cos(\frac{5\pi}{6})=-\frac{\sqrt{3}}{2}\), \(\sin(\frac{5\pi}{6})=\frac{1}{2}\).
• For point \(L\): \(\theta=\frac{7\pi}{6}\), \(\cos(\frac{7\pi}{6})=-\frac{\sqrt{3}}{2}\), \(\sin(\frac{7\pi}{6})=-\frac{1}{2}\).
• For point \(M\): \(\theta=\frac{3\pi}{2}\), \(\cos(\frac{3\pi}{2}) = 0\), \(\sin(\frac{3\pi}{2})=-1\).
• For point \(N\): \(\theta=\frac{5\pi}{4}\), \(\cos(\frac{5\pi}{4})=-\frac{\sqrt{2}}{2}\), \(\sin(\frac{5\pi}{4})=-\frac{\sqrt{2}}{2}\).
• For point \(O\): \(\theta=\frac{7\pi}{12}\), \(\cos(\frac{7\pi}{12})=-\frac{\sqrt{6}-\sqrt{2}}{4}\approx - 0.259\), \(\sin(\frac{7\pi}{12})=\frac{\sqrt{6}+\sqrt{2}}{4}\approx0.966\).
• For point \(P\): \(\theta=\frac{11\pi}{12}\), \(\cos(\frac{11\pi}{12})=-\frac{\sqrt{6}+\sqrt{2}}{4}\approx - 0.966\), \(\sin(\frac{11\pi}{12})=\frac{\sqrt{6}-\sqrt{2}}{4}\approx0.259\).

Answer:

A: \((1,0)\)
B: \((\frac{\sqrt{6}+\sqrt{2}}{4},\frac{\sqrt{6}-\sqrt{2}}{4})\)
C: \((\frac{\sqrt{3}}{2},\frac{1}{2})\)
D: \((\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})\)
E: \((0,1)\)
F: \((\frac{\sqrt{6}-\sqrt{2}}{4},\frac{\sqrt{6}+\sqrt{2}}{4})\)
G: \((\frac{1}{2},\frac{\sqrt{3}}{2})\)
H: \((-\frac{1}{2},\frac{\sqrt{3}}{2})\)
I: \((-1,0)\)
J: \((-\frac{1}{2},-\frac{\sqrt{3}}{2})\)
K: \((-\frac{\sqrt{3}}{2},\frac{1}{2})\)
L: \((-\frac{\sqrt{3}}{2},-\frac{1}{2})\)
M: \((0, - 1)\)
N: \((-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2})\)
O: \((-\frac{\sqrt{6}-\sqrt{2}}{4},\frac{\sqrt{6}+\sqrt{2}}{4})\)
P: \((-\frac{\sqrt{6}+\sqrt{2}}{4},\frac{\sqrt{6}-\sqrt{2}}{4})\)