How do you use the product to sum formulas to write #sin(x+y)cos(x-y)# as a sum or difference?

1 Answer
Feb 14, 2017

#sin(x+y)cos(x-y)=1/2(sin2x+sin2y)#

Explanation:

As #cosA=sin(pi/2-A)#

#sin(x+y)cos(x-y)#

= #sin(x+y)sin(pi/2-(x-y))#

= #sin(x+y)sin(pi/2-x+y)#

Now as #2sinAsinB=cos(A-B)-cos(A+B)#

#sin(x+y)sin(pi/2-x+y)#

= #1/2[cos(x+y-(pi/2-x+y))-cos(x+y+(pi/2-x+y))]#

= #1/2[cos(x+y-pi/2+x-y)-cos(x+y+pi/2-x+y)]#

= #1/2[cos(2x-pi/2)-cos(2y+pi/2)]#

= #1/2[cos(pi/2-2x)-cos(pi/2+2y)]#

  • as #cos(-A)=cosA#

= #1/2(sin2x+1/2sin2y)#

  • as #cos(pi/2-A)=sinA# and #cos(pi/2+A)=-sinA#