0
$\begingroup$

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

$\endgroup$
7
  • $\begingroup$ Have you tried using complex numbers? $\endgroup$ Jan 15 '21 at 10:30
  • $\begingroup$ How i çan use them to solve this ? $\endgroup$ Jan 15 '21 at 10:32
  • $\begingroup$ Consider the similar series but with $\sin$'s on the numerator. $\endgroup$ Jan 15 '21 at 10:33
  • $\begingroup$ Can you explain more, i'm unable to solve it $\endgroup$ Jan 15 '21 at 10:34
  • $\begingroup$ No probelm, take your time $\endgroup$ Jan 15 '21 at 10:38
7
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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 $[\cos(x)]^{n+1}$ $$\dfrac{\cos^{n+1}(x)-e^{ix}\cos^{n+1}(x)}{\cos^{n+1}(x)-e^{ix}\cos^{n}(x)}$$ $\endgroup$ Feb 10 '21 at 3:08
  • $\begingroup$ Also the sum of Raffaele is different, how come? $$\dfrac{\sin(n+1)x}{\sin(x)\cos^{n}(x)}$$ $\endgroup$ Feb 10 '21 at 3:08
  • $\begingroup$ I wish to ask you what does it mean to equate the real and imaginary part? $\endgroup$ Feb 10 '21 at 3:25
  • $\begingroup$ @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) $\endgroup$ Feb 10 '21 at 9:01
  • $\begingroup$ @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. $\endgroup$ Feb 10 '21 at 9:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.