# Prove sin(z1+z2)= sin(z1)cos(z2)+sin(z2)cos(z1), why this way?

1. Mar 18, 2013

### nate9228

1. The problem statement, all variables and given/known data
What I had to prove was sin(z1+z2)= sin(z1)cos(z2)+sin(z2)cos(z1). I did this simply using the e function definitions for sin and cos and it turned out fine. I then flipped to the back of my book to double check and they have a completely different method for proving it. The book is Bak and Newman's Complex analysis.

2. Relevant equations

3. The attempt at a solution
The books method is very non-intuitive, at least to me. It says, "Let z2 be a fixed real number. Then f(z)= sin(z+z2) and g(z)= sin(z)cos(z2)+sin(z2)cos(z) are two entire functions of z which agree for all real values z= z1 and, hence, for all complex values z= z1, as well. Let z=z1 be any such complex number. Then, f(z)= sin(z1+z) and g(z)= sin(z1)cos(z)+ sin(z)cos(z1) agree for all real values z=z2 and, hence, for all complex values z=z2 as well." I do not see how this proves the identity in any way really. Can someone explain, in detail, how this works? And why it would be proved this way when it only takes two minutes to do it using the e function definitions?

2. Mar 19, 2013

### clamtrox

I guess the proof assumes that the identity is true for real numbers, and uses that knowledge to prove it for complex numbers as well.

I agree, it's very silly way of doing it.