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November 24, 2014

November 24, 2014

Posted by **Jaime** on Friday, September 2, 2011 at 8:42pm.

1)using the "x-y-r" definition

2) using identities only

- Geometryy -
**Ms. Sue**, Friday, September 2, 2011 at 8:57pmTimmy, Ryland, Cobra, Jaime -- you must be having an identity crisis tonight.

Please use the same name for all of your posts.

- Geometryy -
**Reiny**, Friday, September 2, 2011 at 9:13pmgiven sec Ø = √13/3 ---> cosØ = 3/√13

also cscØ < 0 ---> sinØ < 0

Ø must be in quadr IV

so x = 3 , r = √13

x^2 + y^2 = r^2

9 + y^2 = 13

y^2 = 4

y = ± 2 , but in IV y = -2

sin Ø = -2/√13

2) you should know that

sin^2Ø + cos^2Ø = 1

sin^2Ø + 9/13 = 1

sin^2Ø = 4/13

sinØ = 2/√13 , but we know that Ø is in IV, where the sine is negative, so

sinØ = -2/√13

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