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October 1, 2014

October 1, 2014

Posted by **Purplechick4** on Wednesday, February 8, 2012 at 12:56am.

- Trig -
**Bosnian**, Wednesday, February 8, 2012 at 1:09ampi < x < 3 pi / 2

180 ° < x < 270 °

That is Quadran III

In Quadran III cosine are negative.

cos ( x ) = + OR - 1 /sqrt [ 1 + tan( x ) ^ 2 ]

In this case :

cos ( x ) = - 1 / sqrt [ 1 + tan( x ) ^ 2 ]

cos ( x ) = - 1 / sqrt [ 1 + ( 9 / 4 ) ^ 2 ]

cos ( x ) = - 1 / sqrt ( 1 + 81 / 16 )

cos ( x ) = - 1 / sqrt ( 16 / 16 + 81 / 16 )

cos ( x ) = - 1 / sqrt ( 97 / 16 )

cos ( x ) = - 1 / [ sqrt ( 97 ) / 4 ]

cos ( x ) = - 4 / sqrt ( 97 )

- Trig -
**Purplechick4**, Wednesday, February 8, 2012 at 1:14amThanks(:

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