How do you find cos(x+y) if sinx= 8/17 in the 1st Quadrant and cosy= 3/5 in the 4th Quadrant?

1 Answer
May 18, 2015

It will help to first calculate #cosx# and #siny#.

#cos^2x = 1-sin^2x = 1-8^2/17^2 = (17^2 - 8^2)/17^2#

#= (289-64)/17^2 = 225/17^2 = 15^2/17^2#

Hence #cosx = +-sqrt(15^2/17^2) = +-15/17#

Since #x# is in the first quadrant, #cosx > 0# so we need to choose the positive square root here:

#cosx = 15/17#

#sin^2y = 1-cos^2y = 1-3^2/5^2 = (5^2-3^2)/5^2#
#= (25-9)/5^2 = 16/5^2 = 4^2/5^2#

Hence #siny = +- sqrt(4^2/5^2) = +-4/5#

Since #y# is in the 4th quadrant, #siny < 0#, so we need to choose the negative square root here:

#siny = -4/5#

Now we can find #cos(x+y)# using the standard formula as follows:

#cos(x+y) = cosxcosy-sinxsiny#

#=(15/17)(3/5)-(8/17)(-4/5)#

#=(15xx3)/(17xx5)+(8xx4)/(17xx5)#

#=((15xx3)+(8xx4))/(17xx5)#

#=(45+32)/85#

#=77/85#