- #1

fatpotato

- Homework Statement
- Calculate the sums ##S_5 = \sum_{k=0}^{n} \frac{\cos(kx)}{\cos^k(x)}## and ##

S_6 = \sum_{k=0}^{n} \frac{\sin(kx)}{\sin^k(x)}##.

- Relevant Equations
- From ##S = \sum_{k=0}^{n} e^{i(a+kb)}## it is possible to derive expressions for the finite sum of sine and cosine where the argument of those functions are of the form##(a+kb)## where ##k## is an integer : $$S_R = \sum_{k=0}^{n} \cos(a + bk) = \cos(a + \frac{nb}{2}) \frac{\sin(\frac{(n+1)b}{2})}{\sin(\frac{b}{2})}$$

And $$S_I = \sum_{k=0}^{n} \sin(a + bk) = \sin(a + \frac{nb}{2}) \frac{\sin(\frac{(n+1)b}{2})}{\sin(\frac{b}{2})}$$

Solution to the problem tells us that ##S_5 + i S_6## is the sum of the terms of a geometric sequence and thus the solutions should be :

$$S_5 = \frac{\sin( (n+1) x)}{\cos^n(x) \sin(x)},\,\,\,\, S_6 = \frac{\cos^{n+1}(x) - \cos((n+1)x)}{\cos^n(x) \sin(x)} , x \notin \frac{\pi}{2} \mathbb{Z}$$

I'm failing to see how they got to the answer. So far, I have written the terms of ##S_5 + i S_6## in both cartesian and polar form to try and find the common ratio ##r## but this method yields nothing.

The other method I fiddled with was taking the result from ##S_R##, taking the special case where ##a=0## and dividing by ##\cos(x)## to try to get a new expression to apply the formula of geometric sums :

$$\frac{S_{R}}{\cos^k(x)} = \sum_{k=0}^{n} \frac{\cos(bk)}{\cos^k(x)} = \cos(\frac{nb}{2}) \frac{\sin(\frac{(n+1)b}{2})}{\sin(\frac{b}{2})} \cdot a_0 \frac{1 - r^{n+1}}{1 - r}$$

Where the first term is ##a_0 = 1## and the common ratio is ##r = \cos^{-1}(x)##, but again, this leads me nowhere. Furthermore, computing series is not my forte and I'm almost certain the last step with the division isn't correct, i.e., it is wrong to embed an expression in a series.

Any help would be appreciated. Don't hesitate to point my mistakes, thank you.

Edit: Latex formatting.

$$S_5 = \frac{\sin( (n+1) x)}{\cos^n(x) \sin(x)},\,\,\,\, S_6 = \frac{\cos^{n+1}(x) - \cos((n+1)x)}{\cos^n(x) \sin(x)} , x \notin \frac{\pi}{2} \mathbb{Z}$$

I'm failing to see how they got to the answer. So far, I have written the terms of ##S_5 + i S_6## in both cartesian and polar form to try and find the common ratio ##r## but this method yields nothing.

The other method I fiddled with was taking the result from ##S_R##, taking the special case where ##a=0## and dividing by ##\cos(x)## to try to get a new expression to apply the formula of geometric sums :

$$\frac{S_{R}}{\cos^k(x)} = \sum_{k=0}^{n} \frac{\cos(bk)}{\cos^k(x)} = \cos(\frac{nb}{2}) \frac{\sin(\frac{(n+1)b}{2})}{\sin(\frac{b}{2})} \cdot a_0 \frac{1 - r^{n+1}}{1 - r}$$

Where the first term is ##a_0 = 1## and the common ratio is ##r = \cos^{-1}(x)##, but again, this leads me nowhere. Furthermore, computing series is not my forte and I'm almost certain the last step with the division isn't correct, i.e., it is wrong to embed an expression in a series.

Any help would be appreciated. Don't hesitate to point my mistakes, thank you.

Edit: Latex formatting.