# Find the missing coordinates.

Fill in the missing coordinates on the unit circle, represented by the letters.

Using sin and cos, we have a $\sin(45^\circ)$ of $\frac{\sqrt{2}}{2}$ and a $\cos(45^\circ)$ of $\frac{\sqrt{2}}{2}$, and a $\cos(60^\circ)$ of $\frac{1}{2}$, and a $\sin(60^\circ)$ of $\frac{\sqrt{3}}{2}$. The coordinates I need are represented by A, B,C, D, E and F.

The answer is A: $\frac{\sqrt{2}}{2}$, $\frac{\sqrt{2}}{2}$, B: $-\frac{1}{2}$, $\frac{\sqrt{3}}{2}$, C: $-\frac{\sqrt{3}}{2}$, $\frac{1}{2}$, D: $-\frac{\sqrt{2}}{2}$, $-\frac{\sqrt{2}}{2}$; E: $\frac{1}{2}, \frac{\sqrt{3}}{2}$; F: $\frac{\sqrt{3}}{2}$, $-\frac{1}{2}$.

How does one come up with these coordinates? Thank you.

• I edited my question. Commented Dec 17, 2016 at 19:26
• Please, have a look at en.wikipedia.org/wiki/Sine#/media/File:Unit_circle_angles.svg to correct your angles. Now, use symmetry. The second coordinate of B is the same as the second coordinate of $60º$ and the first one is minus the first one of $60º.$ In other words $\sin 120=\sin(180-60)=-\sin 60$ and $\cos 120=\cos(180-60)=\cos 60.$
– mfl
Commented Dec 17, 2016 at 19:34
• In each quadrant is first 30 degree then 15 degree again 15 and then 30. Commented Dec 17, 2016 at 19:54
• Actually, the second answer to B is the first answer to $60^\circ$. Commented Dec 17, 2016 at 20:06
• You got it? Or having any doubt? Commented Dec 18, 2016 at 3:04

Point represented as (value of cos, value of sin)

Point A represent $45^\circ$

$sin45^\circ = \frac{1}{\sqrt2}$,

$cos45^\circ = \frac{1}{\sqrt2}$,

Point B represent $120^\circ$

$sin120^\circ = \frac{\sqrt3}{2}$,

$cos120^\circ = -\frac{1}{\sqrt2}$, [ because in second quadrant]

Point C represent $150^\circ$

$sin150^\circ = \frac{1}{2}$,

$cos 150^\circ = -\frac{\sqrt3}{2}$

Point D is in mid of third quadrant so its like $45^\circ$ but in actual its $225^\circ$.

$sin225^\circ = -\frac{1}{\sqrt2}$, [ because in third quadrant]

$cos225^\circ = -\frac{1}{\sqrt2}$, [ because in third quadrant]

Point E represent $300^\circ$

$sin 300^\circ = \frac{\sqrt3}{2}$,

$cos 300^\circ = \frac{1}{2}$

Point F represent $330^\circ$

$sin 330^\circ = \frac{1}{2}$,

$cos 330^\circ = \frac{\sqrt3}{2}$

• These are the same answers I get with a calculator, but the book differs. Commented Dec 17, 2016 at 19:33
• I try to clarify it. Commented Dec 17, 2016 at 19:36
• If still have any doubt you can ask me. Mine pleasure to help you. Commented Dec 17, 2016 at 19:55