# Thread: A small problem on cos a + cos 2a + cos 3a equation

1. ## A small problem on cos a + cos 2a + cos 3a equation

Hello every one,

$cos a + cos2a + cos 3a = \frac {sin (\frac{3a}{2})cos(\frac {3+1}{2}a)}{sin \frac {a}{2}}$

I am a bit embarrased to ask this cause I can go deep for hours in a real calculus problem and enjoy it, but unfortunately is not the same for trigonometry.

I just hate it, I find it so uninteresting and boring. I have looked at some tutorials already.. I always try to do my best before I post but this.. I cannot handle trigonometry that easy... call it psychological what ever...

Once in a while some exercise is asking for it and here I am unable to proove this equality.

Thank you all!

2. ## Re: A small problem on cos a + cos 2a + cos 3a equation

Why don't you write $\cos \left(\frac{1+3}{2}a\right)=\cos(2a)$? Or does it has to be something else? ...

3. ## Re: A small problem on cos a + cos 2a + cos 3a equation

Hello Sirion,

sounds interesting, it might work.. I took the equation from the general term of an infinite series from a solved exercise in my text book.

In addition I have found some tutorials on the internet that mention this identity:

$cos \; a = e^{ix} (the \; real \; part \; of \; it)$

and from a first view this also must be a good way to follow too.

Thank you very much,
bye!

4. ## Re: A small problem on cos a + cos 2a + cos 3a equation

$\cos(a)$ is indeed the real part of Euler's formula $e^{ia}$. I don't tried somehting with Euler's formula but maybe it's possible. For instance you can write: $\cos(x)=\frac{e^{-ix}+e^{ix}}{2}$ ...

Another method where I thought about is by induction.

5. ## Solved!

Oh yes, thank you my friend,
this is what I was not realizing so far, cos (a) is the real part, I only once learned complex numbers in high school and is not easy to think out of real numbers.

I expand this (actually an infinite series, I took n=3 for convinience)

$\frac {e^{(3+1)ix}- 1} { e^(ix) - 1}$

and take only the real part of it (where the real part is cos!!!) and it is okay! So Cool!

6. ## Re: Solved!

Can you show me how far you've done the proof by using Euler's formula?

7. ## Re: A small problem on cos a + cos 2a + cos 3a equation

Finally no success!

If we take

$cos \;a + cos\; 2a + cos \;3a=\sigma_{3}=e^{ia}+e^{2ia}+e^{3ia}$

which means we left out the imaginary part then is also true:

$\sigma_{3}=\frac{e^{n+1}ia - e^{ia}}{e^{ia}}$

as a geometric progression.

Mostly I focused on dominator which is easier. Triggered from your suggestion some experimentation can be done:

$dominator=e^{ia}-1 =\sqrt{e^{ia}}(\sqrt{e^{ia}}-\sqrt{e^{-ia}})=\sqrt{cos\;a+isin\;a}(\sqrt{cos\;a+isin\;a}-\sqrt{\frac{1}{cos\;a+isin\;a}})=cos^{2}a+isin\;a-\frac{cos\;a+isin}{cos\;a+isin}=cos^{2}a+isin\;a-1$

No idea now...however found in a tutorial: $sin\frac{x}{2}=\pm \sqrt{\frac{1-cos\;a}{2}}$

Very tired now...

I spent so much time that already memorized this equation, so no need to generate it on the fly if needed!

Again,
thank you for your time Siron, bye!

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