Adding waves to get resultant wave

[Rococo]

Homework Statement

I need to show that the waves: sin((kx)+(θ/2)) and sin((kx)-(θ/2)), differing in phase by θ, add to give a resultant wave 2sin(kx)cos(θ).
But the answer I get is different so I'm not sure how to do this.

Homework Equations

sinA + sinB = 2sin((A+B)/2)cos((A-B)/2)

The Attempt at a Solution

I tried adding the waves together and so:

A = ((kx)+(θ/2))
B = ((kx)-(θ/2))

(A+B) = ((kx) + (θ/2)) + ((kx)-(θ/2))
(A+B) = (kx) + (θ/2) + (kx) - (θ/2)
(A+B) = (kx) + (kx)
(A+B) = 2(kx)

(A-B) = ((kx)+(θ/2)) - ((kx)-(θ/2))
(A-B) = (kx) + (θ/2) - (kx) + (θ/2)
(A-B) = (θ/2) + (θ/2)
(A-B) = θ

sinA + sinB = 2sin((A+B)/2)cos((A-B)/2)
sin((kx)+(θ/2)) + sin((kx)-(θ/2)) = 2sin((2kx)/2)cos((θ)/2)
sin((kx)+(θ/2)) + sin((kx)-(θ/2)) = 2sin(kx)cos(θ/2)

So I must have have gone wrong somewhere because my final answer is 2sin(kx)cos(θ/2), but it should be 2sin(kx)cos(θ).

 

[ehild]

Your answer is correct.

ehild

 

[Yukoel]

Hello Rococo,
Your solutions is correct.Try θ=90 degrees .R.H.S. is zero right? But the L.H.S ain't. You can check your solutions by substitution.
regards
Yukoel