I'm so close.. trying to prove addition formula for sin(z_1 + z_2)?

Question

Here's the question:

By adding (4) and (5) derive the formula (6)

(4) 2sin(z_1)cos(z_2) = sin(z_1 + z_2) + sin(z_1 - z_2)
(5) 2sin(z_2)cos(z_1) = sin(z_2 + z_1) + sin(z_2 - z_1)

(Those 2 equations are the same except the x_1 and x_2's have been interchanged)

it has to equal

(6) sin(z_1 + z_2) = sin(z_1)cos(z_2) + cos(z_1)sin(z_2)

This is my working out - i am so close by must be missing something, can anyone see where i'm going wrong.

(4) + (5) - i'm just going to type z_1 as z1,z2 etc because there are so many

2sin(z1)cos(z2) + 2sin(z2)cos(z1) =
sin(z1 + z2) + sin(z1-z2) + sin(z2+z1) + sin(z2-z1)

2sin(z1)cos(z2) + 2sin(z2)cos(z1) =
2sin(z1 + z2) + sin(z1-z2) + sin(z2-z1)

Then i did the sub sin(z1) = (e^(iz1) - e^(-iz1))/ 2i

RHS
2(e^(iz1) - e^(-iz1))/ 2i + 2(e^(iz2) - e^(-iz2))/ 2i + (e^(iz1) - e^(-iz1))/ 2i - (e^(iz2) - e^(-iz2))/ 2i + (e^(iz2) - e^(-iz2))/ 2i - e^(iz1) - e^(-iz1))/ 2i


I cant cancel anything?

Answer 1

In the second line of your working sin(z1 + z2) is the same as sin(z2 + z1) so together give 2sin(z1 + z2) which is twice what you want. The sin(z2 - z1) in the same line of working can be replaced by -sin(z1 - z2) as sin(-A) = -sinA for any A. This means that the other two terms in that line cancel one another and you are just left with
2sin(z1)cos(z2) + 2sin(z2)cos(z1) = 2sin(z1 + z2).
Divide by 2 to get the required answer.

I don't know what all the fuss is about. The required answer is just sin(A + B) = sinAcosB + sinBcosA with A = z1 and B = z2.

Answer 2

Back up to
2sin(z1)cos(z2) + 2sin(z2)cos(z1) =
2sin(z1 + z2) + sin(z1-z2) + sin(z2-z1) .

You were almost there... Just notice that sin(z1-z2) and sin(z2-z1) cancel out on the RHS, divide everything by two and you get

→ sin(z1)cos(z2) + sin(z2)cos(z1) = sin(z1 + z2).
Voilà.