Set: Sin/Cos/Tan -> Unit Circle Location

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All 96 terms

TermDefinition
1/2sin of pi/6
root 3/2cos of pi/6
root 3/3tan of pi/6
2csc of pi/6
2 root 3/3sec of pi/6
root 3cot of pi/6
root 2/2sin of pi/4
root 2/2cos of pi/4
1tan of pi/4
root 2csc of pi/4
1cot of pi/4
root 2sec of pi/4
root 3/2sin of pi/3
1/2cos of pi/3
root 3tan of pi/3
2 root 3/3csc of pi/4
root 3/3cot of pi/4
2sec of pi/4
1sin of pi/2
0cos of pi/2
undefinedtan of pi/2
1csc of pi/2
undefinedsec of pi/2
0cot of pi/2
1/2sin of 2pi/3
-root 3/2cos of 2pi/3
-root 3/3tan of 2pi/3
2csc of 2pi/3
-2 root 3/3sec of 2pi/3
- root 3cot of 2pi/3
root 2/2sin of 3pi/4
-root 2/2cos of 3pi/4
-1tan of 3pi/4
root 2csc of 3pi/4
-root 2sec of 3pi/4
-1cot of 3pi/4
root 3/2sin of 5pi/6
-1/2cos of 5pi/6
-root 3tan of 5pi/6
2 root 3/3csc of 5pi/6
-2sec of 5pi/6
-root 3/3cot of 5pi/6
0sin of pi
-1cos of pi
0tan of pi
undefinedcsc of pi
-1sec of pi
undefinedcot of pi
-1/2sin of 7pi/6
-root 3/2cos of 7pi/6
root 3/3tan of 7pi/6
-2csc of 7pi/6
-2 root 3/3sec of 7pi/6
root 3cot of 7pi/6
-root 2/2sin of 5pi/4
-root 2/2cos of 5pi/4
1tan of 5pi/4
-root 2csc of 5pi/4
-root 2sec of 5pi/4
1cot of 5pi/4
-root 3/2sin of 4pi/3
-1/2cos of 4pi/3
root 3tan of 4pi/3
-2 root 3/3csc of 4pi/3
-2sec of 4pi/3
root 3/3cot of 4pi/3
-1sin of 3pi/2
0cos of 3pi/2
undefinedtan of 3pi/2
-1csc of 3 pi/2
undefinedsec of 3 pi/2
0cot of 3 pi/2
-1/2sin of 5pi/3
root 3/2cos of 5pi/3
-root 3/3tan of 5pi/3
-2csc of 5pi/3
2 root 3/3sec of 5pi/3
-root 3cot of 5pi/3
-root 2/2sin of 7pi/4
root 2/2cos of 7pi/4
-1tan of 7pi/4
-root 2csc of 7pi/4
root 2sec of 7pi/4
-1cot of 7pi/4
-root 3/2sin of 11pi/6
1/2cos of 11pi/6
-root 3tan of 11pi/6
-2 root 3/3csc of 11pi/6
2sec of 11pi/6
-root 3/3cot of 11pi/6
0sin of 2pi
1cos of 2pi
0tan of 2pi
undefinedcsc of 2pi
0sec of 2pi
undefinedcot of 2pi

Set Information

Terms 96
Creator myerteaching
Created August 1, 2009
Group lps mentor students
Subjects None
Access Anyone
Edit Creator Only
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Description

Memorize the Unit Circle! Now, when prompted with a sin/cos/tan statement in radians, you should be able to give the coordinates.

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