Posted by Jaime on Friday, September 2, 2011 at 8:42pm.
Timmy, Ryland, Cobra, Jaime -- you must be having an identity crisis tonight.
Please use the same name for all of your posts.
given 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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