Telescopic Series (Trigonometry)

[Heisenberg7]

Homework Statement

Evaluate the sum, $$S_n = 1 + \frac{\cos{x}}{\cos{x}} + \frac{\cos{2x}}{\cos^{2}{x}} + ... + \frac{\cos{nx}}{\cos^{n}{x}}$$ where $n$ is a natural number.

Relevant Equations

below

Greetings,

I would like to gain some insight when it comes to dealing with this problem. Personally, I wasn't able to solve it. I had to look for the solution in the book. I can just tell you that it's a telescopic series, the rest would be too much.

You don't have to post the whole solution, just how you'd approach the problem(like a hint).

 

[fresh_42]

I would use
\begin{align*}
\cos(nx)&=\cos^n(x)-\binom{n}{2}\sin^2(x)\cos^{n-2}(x)+\binom{n}{4}\sin^4(x)\cos^{n-4}(x)\mp\ldots \\&= \sum_{k=0}^{\lfloor n/2 \rfloor}(-1)^k\binom{n}{2k}\sin^{2k}(x)\cos^{n-2k}(x)\\[6pt]
\dfrac{\cos(nx)}{\cos^n(x)}&=\sum_{k=0}^{\lfloor n/2 \rfloor}(-1)^k\binom{n}{2k}\tan^{2k}(x)
\end{align*}

 

[Heisenberg7]

I would use
\begin{align*}
\cos(nx)&=\cos^n(x)-\binom{n}{2}\sin^2(x)\cos^{n-2}(x)+\binom{n}{4}\sin^4(x)\cos^{n-4}(x)\mp\ldots \\&= \sum_{k=0}^{\lfloor n/2 \rfloor}(-1)^k\binom{n}{2k}\sin^{2k}(x)\cos^{n-2k}(x)\\[6pt]
\dfrac{\cos(nx)}{\cos^n(x)}&=\sum_{k=0}^{\lfloor n/2 \rfloor}(-1)^k\binom{n}{2k}\tan^{2k}(x)
\end{align*}

Thank you for you response. This is definitely WAY above my level of understanding. This is listed as a high school level problem, so I expected some high school stuff, but it turns out this could involve some advanced mathematics. Pretty interesting!

 

[Heisenberg7]

It was probably a mistake to post it here instead of posting it in the precalc section.

 

[PeroK]

Homework Statement: Evaluate the sum, $$S_n = 1 + \frac{\cos{x}}{\cos{x}} + \frac{cos{2x}}{\cos^{2}{x}} + ... + \frac{\cos{nx}}{\cos^{n}{x}}$$ where $n$ is a natural number.
Relevant Equations: below

Greetings,

I would like to gain some insight when it comes to dealing with this problem. Personally, I wasn't able to solve it. I had to look for the solution in the book. I can just tell you that it's a telescopic series, the rest would be too much.

You don't have to post the whole solution, just how you'd approach the problem(like a hint).

I would have used the Euler identity:
$$\cos(nx) = Re\big[e^{inx}\big] = Re\big[(\cos x + i\sin x)^n\big]$$Hence:
$$a_n = \frac{\cos(nx)}{\cos^n x} = Re\big[(1 + i\tan x)^n\big]$$This leads to a complex, geometric series with common multiple $(1 + i\tan x)$. Hence:
$$S_n = Re\bigg[\frac{(1 + i\tan x)^{n+1}-1}{i\tan x}\bigg]$$Whether that helps is another matter!

 

[fresh_42]

Thank you for you response. This is definitely WAY above my level of understanding. This is listed as a high school level problem, so I expected some high school stuff, but it turns out this could involve some advanced mathematics. Pretty interesting!

It is only the notation that looks complicated here. The difficulty is that we need to resolve $\cos(nx)$ somehow. The formula I used says
$$
\dfrac{\cos(nx)}{\cos^n(x)}=1-\dfrac{n\cdot(n-1)}{1\cdot 2}\tan^2(x)+\dfrac{n(n-1)(n-2)(n-3)}{1\cdot 2\cdot 3 \cdot 4}\tan^4(x)\mp\ldots
$$
continuing by steps of two. It can formally be proven by induction, or if you like by calculating (or looking up) the first values of $n$ and then writing etc. This results in a telescope sum you mentioned. Summing over all $n$ can be or not be another problem.

 

[PeroK]

It does help, because we can use Euler's formula again:
$$(1 + i\tan x)^{n+1} = \frac{(\cos x + i\sin x)^{n+1}}{\cos^{n+1}x} = \frac{\cos(n+1)x + i\sin(n+1)x}{\cos^{n+1}x}$$And:
$$S_n = Im\bigg[\frac{(1 + i\tan x)^{n+1}-1}{\tan x}\bigg] = \frac{\sin(n+1)x}{\tan x\cos^{n+1}x}$$And, finally:
$$S_n = \frac{\sin(n+1)x}{\sin x\cos^{n}x}$$Advanced high school mathematics!

 

[Heisenberg7]

Thank you, @PeroK and @fresh_42 for trying to help! I would like to present the solution from my book. Here it is:
Notice: $$\frac{\cos{kx}}{\cos^{k}{x}} = \frac{\cos{kx}\sin{x}}{\cos^{k}{x}\sin{x}} = \frac{(\sin{kx}\cos{x} + \cos{kx}\sin{x}) - \sin{kx}\cos{x}}{\cos^{k}{x}\sin{x}} = \frac{\sin{(k+1)x}}{\cos^{k}{x}\sin{x}} - \frac{\sin{kx}}{\cos^{k-1}{x}\sin{x}}$$ By pulling out $\frac{1}{\sin{x}}$ you can clearly see that this is a telescopic series. By using this result in our sum, we get $$S_n = \frac{\sin(n+1)x}{\sin x\cos^{n}x}$$
It's quite elegant really. But you must play with $\frac{\cos{kx}}{\cos^{k}{x}}$ for quite a while to get to the result.

 

[PeroK]

Thank you, @PeroK and @fresh_42 for trying to help! I would like to present the solution from my book. Here it is:
Notice: $$\frac{\cos{kx}}{\cos^{k}{x}} = \frac{\cos{kx}\sin{x}}{\cos^{k}{x}\sin{x}} = \frac{(\sin{kx}\cos{x} + \cos{kx}\sin{x}) - \sin{kx}\cos{x}}{\cos^{k}{x}\sin{x}} = \frac{\sin{(k+1)x}}{\cos^{k}{x}\sin{x}} - \frac{\sin{kx}}{\cos^{k-1}{x}\sin{x}}$$ By pulling out $\frac{1}{\sin{x}}$ you can clearly see that this is a telescopic series. By using this result in our sum, we get $$S_n = \frac{\sin(n+1)x}{\sin x\cos^{n}x}$$
It's quite elegant really. But you must play with $\frac{\cos{kx}}{\cos^{k}{x}}$ for quite a while to get to the result.

It's more obvious once you know the answer to work backwards and see the sequence of steps! I wouldn't beat yourself up if you didn't find that.

 

[Heisenberg7]

It's more obvious once you know the answer to work backwards and see the sequence of steps! I wouldn't beat yourself up if you didn't find that.

Indeed. But, I simply can't help but wonder how an average high school student would find this if they hadn't played with the fraction for at least an hour(religiously), without knowing the result. I worked on this for quite a while, but still couldn't find the result. I felt kind of disappointed, but I guess not everyone is born smart. I know a lot of it is a matter of practice, but believe me, I have practiced trig for a really long time. I suppose I just haven't given it enough time.

 

[PeroK]

Indeed. But, I simply can't help but wonder how an average high school student would find this if they hadn't played with the fraction for at least an hour(religiously), without knowing the result. I worked on this for quite a while, but still couldn't find the result. I felt kind of disappointed, but I guess not everyone is born smart. I know a lot of it is a matter of practice, but believe me, I have practiced trig for a really long time. I suppose I just haven't given it enough time.

This is what makes things harder when you don't know the answer. There's no magic formula. All ideas have a certain specificity. Here, you might have tried to relate one term to the next. Probably for $n = 3$ and $n = 2$. Can you transform one to another in some way? That could take a while to find. Maybe some people can do these things easily, but I can't and I was always good at maths. I wouldn't worry about not getting this one.

In general, trig identities often succumb to the complex approach. That was the first thing I thought of. That's more about knowledge and experience. If I hadn't thought of Euler's formula, then I would have been disappointed!

 

[PeroK]

PS getting better at these sorts of problem isn't something I would focus on. There's a whole world of mathematics to learn beyond trig identities.

 

[Heisenberg7]

PS getting better at these sorts of problem isn't something I would focus on. There's a whole world of mathematics to learn beyond trig identities.

Thank you for motivating me. Eh, where would this world be without people like you or @fresh_42 who are always there to help others!