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January 31, 2015

January 31, 2015

Posted by **jacara** on Tuesday, June 7, 2011 at 10:26am.

Given: point (-2,-6) on the terminal side of the angle in standard position

sin alpha

cos alpha

tan alpha

sec alpha

csc alpha

cot alpha

- trig -
**Henry**, Wednesday, June 8, 2011 at 2:59pm(0,0), (-2,-6).

X = -2-0 = -2.

Y = -6-0 = -6.

r = sqrt((-2)^2+(-6)^2) = sqrt40 =

sqrt(4*10) = 2sqrt10.

sinA = Y/r = -6/2sqrt10 = -3/sqrt10 =

-3sqrt10/10.

cosA = X/r = -2 / 2sqrt10 =

-1 / sqrt10 = -sqrt10 / 10.

tanA = Y/X = -6/-2 = 3.

cscA = 1/sinA = 10 / -3sqrt10 =

10sqrt10 / -3*10 = 10sqrt10 / -30 =

sqrt10 / -3.

secA = 1/cosA = -3 / sqrt10 =

-3sqrt10 / 10.

cotA = 1/tanA = 1/3.

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