0

In this exercise i have to calculate this sum :

$$S=1+\frac{\cos(x)}{\cos(x)}+\frac{\cos(2x)}{\cos^2(x)}+\ldots+\frac{\cos(nx)}{\cos^n(x)}$$

There is no hint in the exercise. I tried to use trigonometric identities but i didn't find the solution

Raffaele
  • 26,371
Badr Eddine
  • 39

1 Answers1

7

Let $$C = 1 + \frac{\cos(x)}{\cos(x)} + \frac{\cos(2x)}{\cos^2(x)} + \cdots + \frac{\cos(nx)}{\cos^n(x)}$$ $$S = \frac{\sin(x)}{\cos(x)} + \frac{\sin(2x)}{\cos^2(x)} + \cdots + \frac{\sin(nx)}{\cos^n(x)}$$ So $$\begin{align}C+iS&=1+\frac{\cos(x)+i\sin(x)}{\cos(x)}+\frac{\cos(2x)+i\sin(2x)}{\cos^2(x)}+...+\frac{\cos(nx)+i\sin(nx)}{\cos^n(x)} \\&=1+\frac{e^{ix}}{\cos(x)}+\frac{e^{2ix}}{\cos^2(x)}+\cdots+\frac{e^{nix}}{\cos^n(x)}\\ &=1+\left(\frac{e^{ix}}{\cos(x)}\right)+\left(\frac{e^{ix}}{\cos(x)}\right)^2+\cdots+\left(\frac{e^{ix}}{\cos(x)}\right)^n\\ &=\frac{\left(\frac{e^{ix}}{\cos(x)}\right)^{n+1}-1}{\left(\frac{e^{ix}}{\cos(x)}\right)-1}\\ &=\frac{e^{(n+1)ix}-\cos^{n+1}(x)}{\cos^{n+1}(x)}\times\frac{\cos(x)}{e^{ix}-\cos(x)}\\ &=\frac{1}{\cos^n(x)}\left(\frac{e^{(n+1)ix}-\cos^{n+1}(x)}{e^{ix}-\cos(x)}\right)\end{align}$$ Now try simplifying this and equating real and imaginary parts. If you need any more help please don't hesitate to ask.

19aksh
  • 12,768
A-Level Student
  • 6,520
  • I have a question:

    $$=\frac{\left(\frac{e^{ix}}{\cos(x)}\right)^{n+1}-1}{\left(\frac{e^{ix}}{\cos(x)}\right)-1}\$$

    I thought this line should be:

    $$=\frac{1-\left(\frac{e^{ix}}{\cos(x)}\right)^{n+1}}{1-\left(\frac{e^{ix}}{\cos(x)}\right)}\$$

    Because the formula for the sum of geometric series is $\dfrac{1-r^{n+1}}{1-r}$

    Why?:

    $$=\frac{e^{(n+1)ix}-\cos^{n+1}(x)}{\cos^{n+1}(x)}\times\frac{\cos(x)}{e^{ix}-\cos(x)}$$

    I multiply the numerator and denominator for $[\cos(x)]^{n+1}$

    $$\dfrac{\cos^{n+1}(x)-e^{ix}\cos^{n+1}(x)}{\cos^{n+1}(x)-e^{ix}\cos^{n}(x)}$$

     – James Warthington Feb 10 '21 at 03:08
  • Also the sum of Raffaele is different, how come?

    $$\dfrac{\sin(n+1)x}{\sin(x)\cos^{n}(x)}$$

     – James Warthington Feb 10 '21 at 03:08
  • I wish to ask you what does it mean to equate the real and imaginary part? – James Warthington Feb 10 '21 at 03:25
  • @JamesWarthington for your first question: the $2$ formulae are identical; it is like the equality $$\frac{-a}{-b}=\frac{a}{b}$$. For your second point, you made a small error; multiplying the top and bottom by $\cos^{n+1}(x)$ actually yields $$\frac{\cos^{n+1}(x)-e^{(n+1)ix}}{\cos^{n+1}(x)-e^{ix}\cos^{n}(x)}$$ For your third question, Raffaele's result is actually the same as mine, it's just that I left the final parts of rearrangement etc to the OP. I can show you how that works if you want. For your fourth question: (continued below) – A-Level Student Feb 10 '21 at 09:01
  • @JamesWarthington (continued) if I have an equation $$p+iq=a+ib$$ where $p,q,a,b\in\mathbb{R}$, then we can deduce that $p=a$ and $q=b$, as real numbers don't change the imaginary component of a complex number, and vice versa. This is called equating real and imaginary parts, as we are equating the real part of each side and also equating the imaginary part of each side. – A-Level Student Feb 10 '21 at 09:05