trig
posted by tony .
Find cos(A+B).
cos A=1/3 and sin B=1/2, with A in quadrant I and B in quadrant IV.

cos 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 
what'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