# Alternate derivation for Cos[X]Cos[Y] - Sin[Y]Sin[X]

Discussion in 'Math' started by kdillinger, Sep 18, 2010.

1. ### kdillinger Thread Starter Active Member

Jul 26, 2009
141
3
I am reviewing complex numbers and how to convert between rectangular, trigonometric, and polar forms. Easy stuff...

What I am trying to figure out is why the angles are simply added in trig and polar forms, so I went back to some basics.

(a+ib)(c+id) is the same as (cos[a]+isin[a])(cos+isin).

Going through the algebra there, one gets:

(cos[a]cos + i(cos[a]sin+sin[a]cos) - sin[a]sin)

where is the leap from the above to the following:

cos[a+b] + i(sin[a+b])

I assume there is an identity that one "just remembers', but I tend to be anal and would like to derive it from first principles. Or is it that the derivation is so insidious and one better choose to "just remember"?

2. ### Ghar Active Member

Mar 8, 2010
655
72
I don't follow your very first statement... those cannot be the same, you have 'c' and 'd' in one equation and they're not there in the other.

The relationship is this:
$Ae^{jx} = Acos(x) + j Asin(x)$

That's Euler's formula.

And, the relationship between rectangular and polar is:
$a + jb = \sqrt(a^2 + b^2)e^{j \cdot tan^{-1}(\frac{b}{a})}$

As for the trig identities you can derive them by converting to polar form and combining exponents then going back to the cos/sin form.

3. ### kdillinger Thread Starter Active Member

Jul 26, 2009
141
3
It was an attempt to show 2 different complex numbers in rectangular form.

I know. The relationship of polar (trigonometric) form to exponential form.

You mean rectangular to exponential?

I am asking how to derive them.

To make is easy I am trying to find the identities necessary to go from:

(cos[a]cos - sin[a]sin) + i(sin[a]cos + cos[a]sin)

to

cos(a+b) + i(sin(a+b))

All literature says "Oh, well, cos[a]cos - sin[a]sin is cos(a+b)."

Is there something more fundamental than this or no?

4. ### Ghar Active Member

Mar 8, 2010
655
72

I did say how you can derive it...
You can use complex numbers... you can write cosine and sine using a sum of complex exponentials.
You can then multiply these out and eventually convert back to the trigonometric form.

There's always something more fundamental because math identities are derived from first principles or can be proven.

The reason I said polar is because polar and exponential are essentially the same thing... it's like you write an 'e'^j instead of a little angle symbol.

Jul 7, 2009
1,585
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Ghar has given you a valuable hint. You can derive many trig identities from Euler's formula (and deriving the angle sum formulas is about the easiest).

If you want to see a derivation of the sum of angles formula from more basic principles, consult any trig textbook (e.g., Palmer, "Plane and Spherical Trigonometry", 1934, page 108; can be downloaded from the web).

6. ### Georacer Moderator

Nov 25, 2009
5,142
1,266
Nice referrence! That's the method I had in mind too. That's the only one I was taught back in highschool.

7. ### kdillinger Thread Starter Active Member

Jul 26, 2009
141
3
I take it no one else can actually do the derivation. Thanks, I will look elsewhere for help.

8. ### Georacer Moderator

Nov 25, 2009
5,142
1,266
Yes... reading from a book can sometimes be tedious...

But how can reading from a thread be easier? Makes me wonder...

9. ### Ghar Active Member

Mar 8, 2010
655
72
Translation:
"I don't understand what you're saying so you must be wrong"

10. ### kdillinger Thread Starter Active Member

Jul 26, 2009
141
3
I never said your wrong it's that I am not interested in cryptic nonsense.

"What do I need to do?"
"You know what you need to do."

"Where do I need to go?"
"You know where you need to go."

I asked how to get from cos[a]cos - sin[a]sin to cos(a+b). It is not apparent to me why one would use exponential form and then convert back to trigonometric form. That is why I am asking if there are other identities for cos[a]cos and sin[a]sin where they somehow reduce to
cos(a+b).

If I was a student and this was for homework, I could forgive your hand waving nonsense. It would have been nice to see an example rather than your hand waving which is no different than telling me to go Google. I did that already, thanks, but it looks like that will be the better option rather than "Paint the Fence" right Mr. Miyagi?

Imma make this post a demotivational poster called HELP FAIL.

/rant

11. ### Georacer Moderator

Nov 25, 2009
5,142
1,266
kdillinger,

someonesdad gave you an accurate referrence on a free book which can be found online. In this book the proof of the theorem is explained in detail.

However, the course of the proof is long and involves pictures of the unitary circle, and copying all of this stuff, while it is already in an easy-to-read form would be unnecessary.

So why don't you do the extra mile and read the stuff yourself, instead of asking us to copy it here? It's you who is supposed to make the effort to learn, not us.

12. ### Ghar Active Member

Mar 8, 2010
655
72
You really don't know how to ask to help.

I told you the general direction you can go in to derive it the first time. Rather than say "I don't still don't follow" or "could you elaborate" you say "I asked how to derive it"

I then said it with a little bit more detail, since apparently you didn't even notice I said anything about deriving the identity. "you can write cosine and sine using a sum of complex exponentials."

If you don't know what that means, again you could say "How does this help me?"
I could then show you a detailed example... but no, you say "I take it no one else can actually do the derivation."

The fail is totally yours.

"Not apparent to me" means acknowledge the fact and ask another question.

13. ### Rick Martin Active Member

Jun 14, 2009
31
2

Do you still require the proof as to why this is true as I know it but will only post it if you still need it.

14. ### retched AAC Fanatic!

Dec 5, 2009
5,201
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That would be nice. Thank you.

15. ### Rick Martin Active Member

Jun 14, 2009
31
2
Okay, look at the below diagram.

From this you can see that the angle for BOD = α+β and angle AOC = α+β so the chords BD and AC must be equal.

Now if we use the distance formula to find the lengths we get the below equations:

√((cos(α+β)-1)^2 + (sin(α+β)-0)^2) = √((cos α-cos β)^2 + (sin α-(-sin β))^2)

By squaring each side and simplifying we obtain:

(cos(α+β)-1)^2 + sin^2(α+β) = (cos α - cos β)^2 + (sin α + sin β)^2

(cos(α+β))^2 - 2cos(α+β) + 1 + (sin(α+β))^2 = (cos α)^2 - 2 cos α cos β + (cos β)^2 + (sin α)^2 + 2 sin α sin β + (sin β)^2

2 - 2 cos(α+β) = 2 - 2 cos α cos β + 2 sin α sin β

If we now use the identity sin^2 θ + cos^2 θ = 1 once on the left-hand side of the equation and twice on the right hand side, you can simplify and get the final result:

cos (α+β) = cos α cos β - sin α sin β

Unfortunately due to the lack of mathematic characters it is hard to really express this but I hope you can slowly follow it as this is how this is derived.

Mar 6, 2009
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Jun 14, 2009
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18. ### BestFriend Member

Sep 22, 2010
31
0
Here's the alternate forms for Cos[X]Cos[Y] - Sin[Y]Sin[X]:

1/2 e^(-i x-i y)+1/2 e^(i x+i y)
and
cos(x+y)