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Simplify this expression?

  1. Oct 12, 2015 #1
    Can someone please show me, step-by-step, how

    cos(θ)sin(θ+φ)-sinθcos(θ+φ)

    simplifies to sinφ?

    I know I have to use trig identities, and I got to cos2θsinφ-sin2θcosφ but I'm not sure where to go from there.

    Thanks
     
  2. jcsd
  3. Oct 12, 2015 #2

    jedishrfu

    Staff: Mentor

    have you tried the ##sin^2\theta + cos^2\theta = 1## ?
     
  4. Oct 12, 2015 #3
    Yeah, if I substitute for both I get:

    [itex]sin(\phi)-sin^2(\theta)sin(\phi)-cos(\phi)+cos^2(\theta)cos(\phi)[/itex]

    But I should probably only substitute for sin or cos so that gives me:

    [itex]sin(\phi)-sin^2(\theta)sin(\phi)-sin^2(\theta)cos(\phi)[/itex]

    or

    [itex]cos^2(\theta)sin(\phi)-cos(\phi)+cos^2(\theta)cos(\phi)[/itex]

    but I'm still not sure where to go from here.
     
  5. Oct 12, 2015 #4

    Svein

    User Avatar
    Science Advisor

    sin(u+v) = sin(u)cos(v)+sin(v)cos(u). Also sin(-u) = -sin(u).
     
  6. Oct 17, 2015 #5
    Maybe someone else already told you but you can use the next identity

    sin(u+-v) = sin(u)cos(v)+-cos(u)sin(v)

    if oyu compare it with you homework then the answers is:

    sin(θ+φ-θ)=sinφ
     
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