Problem: Simplify sinβ‘(xβy)cosβ‘y+cosβ‘(xβy)sinβ‘y\sin (x-y) \cos y+\cos (x-y) \sin ysin(xβy)cosy+cos(xβy)siny.
Answer Choices:
A. 111
B. sinβ‘x\sin xsinx
C. cosβ‘x\cos xcosx
D. sinβ‘xcosβ‘2y\sin x \cos 2 ysinxcos2y
E. cosβ‘xcosβ‘2y\cos x \cos 2 ycosxcos2y
Solution:
Let w=xβyw=x-yw=xβy. Then the given expression is sinβ‘wcosβ‘y+cosβ‘wsinβ‘y=\sin w \cos y+\cos w \sin y=sinwcosy+coswsiny= sinβ‘(w+y)=sinβ‘x\sin (w+y)=\sin xsin(w+y)=sinx.