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

Answer 1
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#
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