# How do I show that cos(z1 + z2) = cos(z1)cos(z2) - sin(z1)sin(z2)

1. May 8, 2010

### squenshl

How do I show that cos(z1 + z2) = cos(z1)cos(z2) - sin(z1)sin(z2)
I let z1 = a + ib & z2 = c + id
So cos(z1 + z2) = cos(a+ib+c+id) = cos(a+ib)cos(c+id) - sin(a+ib)sin(c+id) = cos(z1)cos(z2) - sin(z1)sin(z2). (From the trig identity).

2. May 8, 2010

### lurflurf

Re: Proof

How one approaches such a proof depends how one has defined sine and cosine. One definition is
sin(z):=[e^(iz)-e^(-iz)]/(2i)
cos(z):=[e^(iz)+e^(-iz)]/2i
from
(e^a)(e^b)=e^(a+b)
your result would easily follow
What you have done is assume you result.

3. May 14, 2010

### GRFrones

Re: Proof

(a+ib) and (c+id) are complex numbers, and, in principle, the trigonometric identities are only proved for real numbers... so doing as you said, you're using complex trigonometric identities to prove complex trigonometric identities.

I think the usual ways to define trigonometric functions in complex plane are: using exponential function or using power series representations... in exponential representation, it's very easy to prove the trig. identities, but you'll have to assume some properties of the exponential function... in series representation, most of the theorems are trivially proved for the complex, so you would have to assume less results, but it's a lot harder.