How do you simplify #sin(x+y)+tan(x-y)*cos(x+y)# to trigonometric functions of x and y?
1 Answer
Use the following identities:
[1]
#" " tan(x) = sin(x) / cos(x)# [2]
#" " sin(x+y) = sin(x)*cos(y) + cos(x)*sin(y)# [3]
#" " sin(x-y) = sin(x)*cos(y) - cos(x)*sin(y)# [4]
#" " cos(x+y) = cos(x)*cos(y) - sin(x)*sin(y)# [5]
#" " cos(x-y) = cos(x)*cos(y) + sin(x)*sin(y)# [6]
#" " sin^2(x) + cos^2(x) = 1#
Thus, you can transform:
#sin(x+y) + tan(x - y) * cos(x + y)#
#stackrel("[1] ")(=) " " sin(x+y) + sin(x-y) / cos(x-y) * cos(x+y)#
# = " " sin(x+y) + (sin(x-y) cos(x+y))/cos(x-y)#
#stackrel("[2],[3],[4],[5]")(=) sin(x) cos(y) + cos(x) sin(y) + ((sin(x)cos(y) - cos(x) sin(y))*(cos(x)cos(y) - sin(x)sin(y))) / (cos(x)cos(y) + sin(x)sin(y))#
# = " " sin(x) cos(y) + cos(x) sin(y) + (sin(x)cos(x)cos^2(y) - sin(y)cos(y)cos^2(x) - sin(y)cos(y)sin^2(x) + sin (x)cos(x)sin^2(y)) / (cos(x)cos(y) + sin(x)sin(y)) #
# = " " sin(x) cos(y) + cos(x) sin(y) + (color(green)(sin(x)cos(x)cos^2(y)) - sin(y)cos(y)cos^2(x) - sin(y)cos(y)sin^2(x) + color(green)(sin (x)cos(x)sin^2(y))) / (cos(x)cos(y) + sin(x)sin(y)) #
# = " " sin(x) cos(y) + cos(x) sin(y) + ( sin(x)cos(x)(cos^2(y) + sin^2(y)) - sin(y)cos(y)(cos^2(x) + sin^2(x))) / (cos(x)cos(y) + sin(x)sin(y)) #
# stackrel("[6] ")(=)" " sin(x) cos(y) + cos(x) sin(y) + (sin(x)cos(x) - sin(y)cos(y))/(cos(x)cos(y) + sin(x)sin(y))#
Unfortunately, I don't see how you could efficiently simplify this expression any further.
Hope that helped!
