How Does cos(θ)sin(θ+φ)-sinθcos(θ+φ) Simplify to sinφ?

[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

 

[jedishrfu]

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

 

[leaf]

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.

 

[Svein]

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

 

[andres_porras]

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

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