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April 1, 2015

April 1, 2015

Posted by **tony** on Sunday, June 16, 2013 at 11:20pm.

cos A=1/3 and sin B=-1/2, with A in quadrant I and B in quadrant IV.

- trig -
**Bosnian**, Monday, June 17, 2013 at 12:42amcos A = 1 / 3

sin A = + OR - sqrt ( 1 - cos A ^ 2 )

sin A = + OR - sqrt ( 1 - ( 1 / 3 ) ^ 2 )

sin A = + OR - sqrt ( 1 - 1 / 9 )

sin A = + OR - sqrt ( 9 / 9 - 1 / 9 )

sin A = + OR - sqrt ( 8 / 9 )

sin A = + OR - sqrt ( 4 * 2 / 9 )

sin A = + OR - sqrt ( 4 ) * sqrt ( 2 ) / sqrt ( 9 )

sin A = + OR - 2 * sqrt ( 2 ) / 3

sin A = + OR - ( 2 / 3 ) * sqrt ( 2 )

In quadrant I sine are positive so :

sin A = ( 2 / 3 ) * sqrt ( 2 )

sin B = - 1 / 2

sin B = + OR - sqrt ( 1- cos B ^ 2 )

cos B = + OR - sqrt ( 1 - ( - 1 / 2 ) ^ 2 )

cos B = + OR - sqrt ( 1 - 1 / 4 )

cos B = + OR - sqrt ( 4 / 4 - 1 / 4 )

cos B = + OR - sqrt ( 3 / 4 )

cos B = + OR - sqrt ( 3 ) / 2

In quadrant IV cosine are positive so :

cos B = sqrt ( 3 ) / 2

cos ( A + B ) = cos A * cos B - sin A * sin B

cos ( A + B ) = ( 1 / 3 ) * sqrt ( 3 ) / 2 - ( 2 / 3 ) * sqrt ( 2 ) * ( - 1 / 2 )

cos ( A + B ) = ( 1 / 6 )sqrt ( 3 ) + ( 2 / 6 ) * sqrt ( 2 )

cos ( A + B ) = ( 1 / 6 ) [ sqrt ( 3 ) + 2 sqrt ( 2 ) ]

cos ( A + B ) = [ sqrt ( 3 ) + 2 sqrt ( 2 ) ] / 6

- trig -
**Steve**, Monday, June 17, 2013 at 4:30amwhat's all this work?

A is in QI, so sinA = √8/3

B is in QIV so cosB = √3/2

and then as done in the final paragraph

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