Memorizing the Unit Circle: Tips & Tricks

[justPAB]

I am having a real tough time memorizing the unit circle and it's values. What would you suggest to make easier for me to remember the quadrants, square roots, and radians?

 

[arildno]

Remember what is the correct values for 30 degrees and 45 degrees only.

Deduce the rest when you encounter them.

For example, 120 degrees is 30 degrees with respect to the positive y-axis, the x-coordinate negative.

That's all you need.

 

[justPAB]

Thank you! That makes sense, I'll have to memorize them like that; it's been a couple years since trigonometry for me and I missed a semester of math this fall. Onto calculus 2 in spring and I don't want to fail lol.

 

[mathwonk]

just remember that all the way around is 2pi radians, amnd then draw apicture and divide up the circle to get fractions of it. e.g. straight up is 1/4 a circle so 2pi/4 = pi/2 radians.

and remember to start at (1,0) (the unit on the positive, i.e. right hand, x axis) and go counterclockwise around the circle.

 

[Mark44]

The coordinates of the angles in the first quadrant are not too difficult to memorize. The x coordinate is the cosine of the angle, and the y coordinate is the sine of the angle. The coordinates follow a nice pattern.

0 (sqrt(4)/2, sqrt(0)/2) = (1, 0)
pi/6 (sqrt(3)/2, sqrt(1)/2)
pi/4 (sqrt(2)/2, sqrt(2)/2)
pi/3 (sqrt(1)/2, sqrt(3)/2)
pi/2 (sqrt(0)/2, sqrt(4)/2) = (0, 1)

As arildno said, you can deduce the angles in the other quadrants.

 

[HallsofIvy]

$\pi/4$ radians is exactly half way between 0 and 90 so is the line y= x. The unit circle has equation $x^2+ y^2= 1$ so with y= x, [itex]x^2+ x^2= 2x^2= 1[/otex] and $x^2= 1/2$. $x= cos(\pi/4)= \sqrt{1/2}= \sqrt{2}/2$. Of course, $sin(\pi/4)= y= x= \sqrt{2}/2$ also.

For $\pi/3$ and $\pi/6$ think of the equilateral triangle which has angles of $\pi/6$ radians. Dropping a perpendicular from one vertex to the opposite side divides the triangle into two right triangles with angles $\pi/3$ and $\pi/6$. If we take one side of the equilateral triangle to be 2, that will be the hypotenuse of a right triangle and the side opposite the $\pi/6$ angle will be 1. By the Pythagorean theorem, $1^2+ x^2= 2^2$ where x is the length of the other leg (the altitude of the equilateral triangle). That is, $x^2= 4- 1= 3$ so that $x= \sqrt{3}$. $\sin(\pi/6)= 1/2$, $cos(\pi/6)= \sqrt{3}/2$ and $sin(\pi/3)= \sqrt{}/2$, $cos(\pi/3)= 1/2$.

Of course, multiples of $\pi/2$ are on the axes so those should be easy.

 

[romjke]

if you draw a circle and label the x and y-axis you can determine the distance that each point on the unit circle is from the point (0,0)

 

[Mark44]

if you draw a circle and label the x and y-axis you can determine the distance that each point on the unit circle is from the point (0,0)

All points on the unit circle are 1 unit from (0, 0).

 

[physeven]

A way I used to memorize them was to put everything in terms of pi/4 or pi/2. This makes things colossally easier for me. And most of the time I never have to use a calculator for any trig operations.

 

[HallsofIvy]

What did you do about $\pi/3$ and $2\pi/3$?

 

[physeven]

What did you do about $\pi/3$ and $2\pi/3$?

can be put in terms of pi/6.

 

[HallsofIvy]

can be put in terms of pi/6.

Yes, they can. But you said "put everything in terms of pi/4 or pi/2."

 

[physeven]

Yes, they can. But you said "put everything in terms of pi/4 or pi/2."

well i guess what i should have said from the beginning was 'put in terms of pi/2, pi/4 and pi/6.' that would've saved us some reply notifications :P.