More trigonometry help

For this problem, we're asked to proof it, any help?

sin(a+b)cos(b)-cos(a+b)sin(b)=sin(a)

I try to apply the identities and all, then it becomes one mess of distribution and I just cry... Help? :(

we'll give it a shot...

LHS:
sin (a + b) cos b = (sin a cos b + cos a sin b)cos b = sin a cos^2 b + cos a cos b sin b
cos (a + b)sin b = (cos a cos b - sin a sin b)sin b = cos a cos b sin b - sin a sin ^2 b

sin (a + b) cos b - cos (a + b) sin b = sin a cos^2 b + cos a cos b sin b - [cos a cos b sin b - sin a sin ^2 b]
= sin a cos^2 b + cos a cos b sin b - cos a cos b sin b + sin a sin^2 b

the cos a cos b sin b terms cancel, leaving
sin a cos^2 b + sin a sin^2 b
= sin a (cos^2 b + sin^2 b)
and since cos^2 b + sin ^2 b = 1, you have sin a(1) = sin a = RHS QED

sin(a+b)=sina*cosb+sinb*cosa

and cos(a+b)=cosa*cosb-sina*sinb

so substituting
sin(a+b)*cos(b)-cos(a+b)*sin(b)=sin(a)

(sina*cosb+sinb*cosa)*(cosb)-(cosa*cos…
(sina*cosb^2+sinb*cosa*cosb)-(cosa*cos…

now as you can see we have a +sinb*cosa*cosb and a -sinb*cosa*cosb so adding those we have 0 and we are left with
sina*cosb^2-(-sina*sinb^2)

sina*cosb^2+sina*sinb^2=sina now we can factor out a sina
sina*(cosb^2+sinb^2)=sina

now remember that cosb^2+sinb^2=1 so substituting that in we have

sina*(1)=sina

sina=sina

and there you have it
hope that helped

wouldn't it be sin((a+b)-b)? then simplified to sin(a)?