# Trigometric addition of complex numbers!

1. Apr 19, 2007

### Mathman23

Hello

I got some funny idears don't know if they are true, but I will share them with you.

1. The problem statement, all variables and given/known data

I would like to prove these formula
(1)$$sin(z_1 + z_2) = sin(z_1) \cdot cos(z_2) + sin(z_2) \cdot cos(z_1)$$

(2)$$cos(z_1 + z_2) = sin(z_1) \cdot sin(z_2) - cos(z_2) \cdot cos(z_1)$$

(3)$$cos(z)^2 + sin(z)^2 = 1$$

for any complex number $$z_1, z_2, z \in \mathbb{C}$$

3. The attempt at a solution
Proof for (1)

(step a)Let $$z_1 = x_1 + i \cdot y_1$$ and $$z_2 = x_2 + i \cdot y_2$$

(step b)According to the rules of addition of complex numbers

$$z_1 + z_2 = (x_1 + x_2) + i(y_1 + y_2)$$, where $$z_1 + z_2 \in \mathbb{C}$$

(step c)I denote $$x_1 + x_2 = x$$ and $$y_1 + y_2 = y$$

(step d)thus by replacing x and y by their respective polar coordinant representation:

$$z_1 + z_2 = (r \cdot cos(\theta)) + i \cdot ( r \cdot sin(\theta))$$, where $$z_1 + z_2 \in \mathbb{C}$$

(step e)which can be reformulated to:

$$z_1 + z_2 = r \cdot (cos(\theta)) + i \cdot ( sin(\theta))$$.

(step f)Which equals according to the euler identity:

$$z_1 + z_2 = r \cdot e^{i \theta}$$.

(step g)Then I apply sinus on both sides of the equality and get

$$\begin{array}{ccc}sin(z_1 + z_2) = cosh(sin(\theta) \cdot r) \cdot sin(cos(\theta) \cdot r) + sinh(sin(\theta) \cdot r) \cdot cos(cos(\theta) \cdot r) \cdot i \end{array}$$

From the above we see that $$x = r \cdot cos(\theta)$$ and $$y = r \cdot sin(\theta)$$

(step h)
With this in mind we rearrange the above and remembers the identity for x and y.

$$sin(z_1 + z_2) = sin(x_1 + x_2) \cdot cosh(y_1 + y_2) + cos(x_1 + x_2) \cdot sinh(y_1 + y_2) \cdot i$$

I have thus proved the sinus addition formula for complex numbers. (q.e.d)

(2) Proof

The steps are the same until (step g) where I take cosine on both sides of the equality and then obtain

$$cos(z_1 + z_2) = cos(x_1 + x_2) \cdot cosh(y_1 + y_2) - sin(x_1 + x_2) \cdot sinh(y_1 + y_2) \cdot i$$

q.e.d.

Test for (1): Let $$z_1 = 2 + 2i$$ and $$z_2 = 4 + 2i$$

Then by applying my formula on

$$z_1 + z_2 = 6 + 4i$$

$$sin(z_1 + z_2) = sin(6 + 4i) = sin(6) \cdot cosh(4) + cos(6) \cdot sinh(4) \cdot i$$

Thus my formula is true.

Test for (1): Let $$z_1 = 2 + 2i$$ and $$z_2 = 4 + 2i$$

Then by applying my formula on

$$z_1 + z_2 = 6 + 4i$$

$$cos(z_1 + z_2) = cos(6 + 4i) = cos(6) \cdot cosh(4) - sin(6) \cdot sinh(4) \cdot i$$

Thus my formula is true.

Best Regards

Fred

2. Relevant equations

p.s. Does my proof look okay? If yes could somebody please give a hint for (3)??

2. Apr 19, 2007

### Gib Z

I haven't looked through the 2 proofs very well, but they look fine. For 3, Write sin x and cos x in exponential form from Eulers Formula. Write z = a+bi, and expand.

3. Apr 19, 2007

### Mathman23

Hi Gib and Dick,

Regardin (3)

Proof:

Let $$z = a +ib$$ be a complex number.

Then according to Eulers formula:

$$sin(z) = \frac{e^{iz} - e^{-iz}}{2i}$$

$$cos(z) = \frac{e^{iz} + e^{-iz}}{2}$$

Then by squaring the above and adding them together (My motovation for doing this part, is that that the point (sin(z),cos(z)) lies on the unit cicle (x^2 + y^2 = 1) ?)

$$(\frac{e^{iz} - e^{-iz}}{2i})^2 + (\frac{e^{iz} + e^{-iz}}{2})^2 =1$$

Thus the formula for $$sin(z)^2 + cos(z)^2 =1$$

where z is complex number is true. (q.e.d)

Best Regards
Fred

p.s. You you please be so kind to comment my proofs for (1) and (2) :)

p.p.s. Thank You Dick for you correction of my trig.add mistakes. It was a dumb a** mistake on my part.

Last edited: Apr 19, 2007
4. Apr 19, 2007

### Dick

Square the exponential expressions and add them. cos(z) equals negative cos(a+ib)!? Don't be silly.

5. Apr 19, 2007

### Mathman23

Hi Dick,

Please look at my edited post #3.

Best Regards
Fred

6. Apr 19, 2007

### Dick

Looks fine. Did you actually expand the squares and use the rules of exponentials to verify the sum is one?

7. Apr 19, 2007

### Mathman23

No I didn't.

I will just look it up.

Last edited: Apr 19, 2007
8. Apr 19, 2007

### Dick

NO. Like this:

$$(\frac{e^{iz} + e^{-iz}}{2})^2+(\frac{e^{iz} - e^{-iz}}{2i})^2 = ?? = 1$$

Where did that crazy sum come from?

Last edited: Apr 19, 2007
9. Apr 19, 2007

### Mathman23

Dear Dick

It was a cut and past error on my part , it should have said

(sin(z))^2 + (cos(z))^2 = 1.

Best regards

Fred

p.s. You mean by expand the above I write them as the sum of their squared respective power series?

Last edited: Apr 19, 2007
10. Apr 19, 2007

### Dick

No again. (cos(z))^2=((exp(iz)+exp(-iz))/2)^2=exp(2iz)/4+1/2+exp(-2iz)/4. Do you see why? What's sin(z)^2? Add them.

11. Apr 19, 2007

### Mathman23

Hello Dick,

I forgot to be carefull. Sorry.

By repeating you thema I get sin(z)^2

$$sin(z)^2 = (\frac{e^{iz}-e^{-iz}}{2i})^2 = \frac{-(e^{iz})^2}{4} + \frac{1}{2} - \frac{(e^{-iz})^2}{4}$$

Therefore sin(z)^2 + cos(z)^2 = 1.

Best Regards
Fred

Last edited: Apr 19, 2007
12. Apr 19, 2007

### Dick

13. Apr 19, 2007

### Mathman23

I changed my post #11 above.

How does the proof of (3) look now? Am I steppin in the right direction?

Best Regards.

/Fred

Last edited: Apr 19, 2007
14. Apr 19, 2007

### Gib Z

Looks like the proof is complete. Well done, I guess there was no need to let z=a+bi after all.

15. Apr 20, 2007

### Mathman23

Hi Just a clean write of part 3,

Proof (Part 3):

Let $$z$$ be a complex number.

Then according to Eulers formula:

$$sin(z) = \frac{e^{iz} - e^{-iz}}{2i}$$

$$cos(z) = \frac{e^{iz} + e^{-iz}}{2}$$

Its know that

$$(sin(z))^2 = (\frac{e^{iz} - e^{-iz}}{2i})^2 = -\frac{(e^{iz})^2}{4} + \frac{1}{2} - \frac{(e^{-iz})^2}{4}$$

$$(cos(z))^2 = (\frac{e^{iz} + e^{-iz}}{2})^2 = \frac{(e^{iz})^2}{4} + \frac{1}{2} + \frac{(e^{-iz})^2}{4}$$

By addin them together I get

$$(sin(z))^2 + (cos(z))^2 = (-\frac{(e^{iz})^2}{4} + \frac{1}{2} - \frac{(e^{-iz})^2}{4}) + (\frac{(e^{iz})^2}{4} + \frac{1}{2} + \frac{(e^{-iz})^2}{4}) = 1$$

Thus the formula for $$sin(z)^2 + cos(z)^2 = 1$$

where z is complex number is true. (q.e.d)

Best Regards
Fred

p.s. Could somebody please be so kind and look through this part 3, and see if I have overlooked anything?

Last edited: Apr 20, 2007
16. Apr 20, 2007

### Dick

Looks good to me.

17. Apr 21, 2007

### Mathman23

Regarding (1) and (2) are they okay too?

Any need to try to factor my formula result so they look like the original formulas

sin(z1 + z2) = sin(z1) * cos(z2) + sin(z2) * cos(z1) ?

or is my way of formulating it consistant??

Best Regards
Fred

18. Apr 21, 2007

### Dick

(1) and (2) don't look so great. Just do them the same way you did (3). Replace all of the cos's and sin's by exponentials on both sides of the equations. Then multiply everything out and show both sides are equal. BTW you have the two terms on the RHS reversed in your statement of (2).

19. Apr 21, 2007

### Mathman23

Okay thanks,

Thought I was too lucky about 1 and 2.

by using the euler formula for sin(z1+z2)

I get following:

$$\begin{array}{cccc}sin(z_1+z_2) &=& \frac{e^{i \cdot z_1} + e^{- i \cdot z_1}}{2} \cdot \frac{e^{i \cdot z_2} - e^{- i \cdot z_2}}{2i} + \frac{e^{i \cdot z_2} + e^{- i \cdot z_2}}{2} \cdot \frac{e^{i \cdot z_1} - e^{- i \cdot z_1}}{2i} \\ &=& sin(z_1) \cdot cos(z_2) + sin(z_2) \cdot cos(z_1) \end{array}$$

By using euler on cos(z_1 + z_2)

$$\begin{array}{cccc}cos(z_1+z_2) &=& \frac{e^{i \cdot z_2} - e^{- i \cdot z_2}}{2i} \cdot \frac{e^{i \cdot z_1} - e^{- i \cdot z_1}}{2i} + \frac{e^{i \cdot z_2} + e^{- i \cdot z_2}}{2} \cdot \frac{e^{i \cdot z_1} + e^{- i \cdot z_1}}{2} \\ &=& cos(z_1) \cdot cos(z_2) - sin(z_1) \cdot sin(z_2) \end{array}$$

Is that what you mean?

Best regards

Fred

Last edited: Apr 21, 2007
20. Apr 21, 2007

### Gib Z

Yes that Looks excellent. Good work.