Timmy, Ryland, Cobra, Jaime -- you must be having an identity crisis tonight.
Please use the same name for all of your posts.
given that sec theta = radical13/whole number 3, and csctheta<0 find sin theta using the 2different methods;
1)using the "x-y-r" definition
2) using identities only
2 answers
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
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
1 answer