1

The problem is to find the summary of this statement: $$\sin(x) + \sin(3x) + \sin(5x) + \dotsb + \sin(2n - 1)x = $$

I've tried to rewrite all sinuses as complex numbers but it was in vain. I suppose there is much more complicated method to do this. I think it may be solved somehow using complex numbers or progressoins.

How it's solved using complex numbers or even without them if possible?

Thanks a lot.

Em.
  • 15,981
Sergei
  • 67

2 Answers2

9

Note that we wish to evaluate the sum $\sum_{k=1}^n\sin((2k-1)x)$. Then, using the addition angle law, we have

$$\sin(x)\sin((2k-1)x)=\frac12\left(\cos(2(k-1)x)-\cos(2kx)\right)$$

Now, sum the telescoping series and divide by $\sin(x)$.

Mark Viola
  • 179,405
8

Hint:

Your complex sum is: $$S=e^{ix}+e^{3ix}+....e^{(2n-1)ix}=e^{ix}\frac{ 1-e^{2inx } }{ 1-e^{2ix} }$$

then you take the imaginary part.

You will obtain

$$\frac{ \sin^2(nx) }{ \sin(x) }$$

with the condition $x\neq k\pi$ where $k\in\mathbb Z$.

Sebastiano
  • 7,649
hamam_Abdallah
  • 62,951
  • 1
    @Dr.MV This is quite easy as well; the trick is to collect $e^{inx}$ from the numerator and $e^{ix}$ from the denominator. Piece of cake. ;-) – egreg Oct 13 '16 at 20:18
  • @Dr.MV Provided you remember the product-to-sum formula. ;-) I always have to look it up. – egreg Oct 13 '16 at 20:26
  • Furthermore, you can have two sums for the price of one… – Bernard Oct 13 '16 at 20:35
  • @Bernard, do you refer to Dr. MV solution? I would rather word it backwords: One solution to the price two sums. I found that more mathematician like, contra intuitive, but "true" just like: 1 guy enters the house, 2 leave, now 1 has to enter for it to be empty.. – Imago Oct 13 '16 at 20:44
  • @Imago: No, I was alluding to Abdallah Hammam's solution. I love the logic (Lewis Carroll or Marx Brothers?) – Bernard Oct 13 '16 at 20:53