# Simplify this expression?

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1. Oct 12, 2015

### leaf

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. Oct 12, 2015

### Staff: Mentor

have you tried the $sin^2\theta + cos^2\theta = 1$ ?

3. Oct 12, 2015

### leaf

Yeah, if I substitute for both I get:

$sin(\phi)-sin^2(\theta)sin(\phi)-cos(\phi)+cos^2(\theta)cos(\phi)$

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

$sin(\phi)-sin^2(\theta)sin(\phi)-sin^2(\theta)cos(\phi)$

or

$cos^2(\theta)sin(\phi)-cos(\phi)+cos^2(\theta)cos(\phi)$

but I'm still not sure where to go from here.

4. Oct 12, 2015

### Svein

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

5. Oct 17, 2015

### andres_porras

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