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Posted by **Keenia** on Monday, December 12, 2011 at 8:59am.

- trig -
**Reiny**, Monday, December 12, 2011 at 9:22am16π is the same as exactly 8 rotations, so

-cot(π/4 + 16π) = -cot(π/4)

= -1/tan(π/4) = -1/1 = -1

- trig -
**Anonymous**, Monday, December 12, 2011 at 9:36amcot ( A + B ) = 1 / tan ( A + B )

tan ( A + B )= ( tanA + tanB )/( 1 - tanA * tanB )

In this case:

A = pi / 4

B = 16 pi

tan ( pi / 4 ) = tan 45° = 1

tan ( 16 pi )= cot ( 8 * 2 pi ) = tan ( 8 * 360° ) = 0

tan ( pi / 4 + 16 pi ) =

[ ( tan ( pi / 4 ) + tan ( 16 pi ) ]/[ 1 - tan( pi / 4 ) * tan ( 16 pi ) ] =

( 1 + 0 ) / ( 1 - 1 * 0 ) =

1 / ( 1 - 0 ) =

1 / 1 = 1

tan ( pi / 4 + 16 pi ) = 1

cot ( pi / 4 + 16 pi ) =

1 / tan ( pi / 4 + 16 pi ) =

1 / 1 = 1

- cot ( pi / 4 + 16 pi ) = - 1

- trig -
**Anonymous**, Monday, December 12, 2011 at 9:37amtan ( 16 pi )= tan ( 8 * 2 pi ) = tan ( 8 * 360° ) = 0

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