Multiple angles : Reducing the sum

[brotherbobby]

Homework Statement

Show that $\large{\boxed{\tan\theta+2\tan 2\theta+4\tan 4\theta+8 \cot 8\theta = \cot\theta}}$

Relevant Equations

1. $\tan 2\theta = \frac{2\tan\theta}{1-\tan^2\theta}$
2. $\tan 4\theta = \frac{4\tan \theta-4\tan^3 \theta}{1-6\tan^2\theta+\tan^4\theta}$ (borrowed from a problem that asked to prove as much).
3. $\tan 8\theta =?, \cot 8\theta = ?$ (No readymade equations available)

Problem Statement : Let me copy and paste the problem as it appears in the text :

Attempt : I haven't been able to make any significant attempt at solving this problem, am afraid. I tried to reduce all the higher submultiple angles $2\theta, 4\theta, 8\theta$ into $\theta$, but the resulting equation became unwieldy.

Request : As I have failed to do my fair share, the barest of hints to start me off would be welcome.

 

[Office_Shredder]

$tan(8\theta)= tan(2(4\theta))$. You can apply the double angle formula, and then use your known result for the quadruple angle.

$cot(8\theta)=1/tan(8\theta)$ so once you know one you know the other.

I don't know if this is the best way to go about this but at least you can give it a shot now.

 

[Lnewqban]

Please, see:
https://en.wikipedia.org/wiki/List_of_trigonometric_identities

 

[docnet]

Spoiler alert! A solution to this problem is in the link: Two approaches to this problem

 

[SammyS]

@brotherbobby, where are you?

Here is a slightly different hint.

See what you can do with $\displaystyle \tan(\phi) + 2 \cot(2 \phi)$ .

Once we see some workings from you, you may expect some additional suggestions.

 

[vela]

Another way to attack the problem is using complex exponentials. Start with
$$\tan \phi = \frac{1}{i} \frac{e^{i\phi}-e^{-i\phi}}{e^{i\phi}+e^{-i\phi}} = \frac{1}{i} \frac{e^{2i\phi}-1}{e^{2i\phi}+1} = \frac{1}{i} \frac{z-1}{z+1},$$ where $z=e^{2i\phi}$. You can express the other terms in the identity in terms of $z$ as well. Then do a lot of algebra.

 

[brotherbobby]

Hello @Office_Shredder and @SammyS, thank you for your help and sorry for coming in late. I will solve the problem using two methods, neither of them mine. I will use @vela 's method later (post #6 above).

Problem statement : $\large{\boxed{\tan\theta+2\tan 2\theta+4\tan 4\theta+8 \cot 8\theta = \cot\theta}}$.

First Attempt (brute force) : I will paste my solution done using Autodesk Sketchbook$^{\circledR}$. I hope I am not in violation of something.

Second attempt (elegant) : $\text{What is} \tan\theta-\cot\theta$? I will derive an expression for this and then use it to do the problem above.

 

[SammyS]

Hello @Office_Shredder and @SammyS, thank you for your help and sorry for coming in late. I will solve the problem using two methods, neither of them mine. I will use @vela 's method later (post #6 above).

$\tan\theta+2\tan 2\theta+4\tan 4\theta+8 \cot 8\theta = \cot\theta$

Let's look at the following part of your Second Attempt.

Second attempt (elegant) : $\text{What is} \tan\theta-\cot\theta$? I will derive an expression for this and then use it to do the problem above.

So, if you had taken my suggestion in Post #5, the above shows that you could have found that:

$\displaystyle \tan(\phi) + 2 \cot(2 \phi) = \cot( \phi )$.

Then use this result recursively, starting with the last two terms on the left hand side of your original expressions.

Use $4\theta$ in place of $\phi$ and you get:

$\displaystyle 4(\tan(4\theta) + 2 \cot(2 (4\theta) ) )= 4\cot(4\theta)$

Repeat using suitable replacement for $\phi$.

 

[vela]

I will use @vela 's method later (post #6 above).

I'll warn you that it's a lot of tedious algebra. It may not be worth solving the problem this way except perhaps to satisfy your curiosity.

 

[brotherbobby]

$\tan\theta+2\tan 2\theta+4\tan 4\theta+8 \cot 8\theta = \cot\theta$

Let's look at the following part of your Second Attempt.

View attachment 295749
So, if you had taken my suggestion in Post #5, the above shows that you could have found that:

$\displaystyle \tan(\phi) + 2 \cot(2 \phi) = \cot( \phi )$.

Then use this result recursively, starting with the last two terms on the left hand side of your original expressions.

Use $4\theta$ in place of $\phi$ and you get:

$\displaystyle 4(\tan(4\theta) + 2 \cot(2 (4\theta) ) )= 4\cot(4\theta)$

Repeat using suitable replacement for $\phi$.

Yes @SammyS, the method you had suggested is close to the second of my attempts above. Thank you. To be honest, it is @vela's method that interests me at the moment. I'd like to try it presently.

 

[SammyS]

To be honest, it is @vela's method that interests me at the moment. I'd like to try it presently.

As @vela said, it can involve a lot of tedious algebra.
However, using it to show your result (1) may be enough to satisfy your curiosity.

$\displaystyle \cot( \phi ) - \tan(\phi) = 2 \cot(2 \phi) \quad$ (1)

 

[brotherbobby]

I'll warn you that it's a lot of tedious algebra. It may not be worth solving the problem this way except perhaps to satisfy your curiosity.

No worries, we are young people, hungry to know. Curious as hell.

Problem statement : $\large{\boxed{\tan\theta+2\tan 2\theta+4\tan 4\theta+8 \cot 8\theta = \cot\theta}}$.

Third Attempt (using complex numbers) ​

​

[Prelude to complex numbers : I remember from school days that the complex number $z$ can be written in the "rectangular" and "polar" forms in the following way : $z=x+iy=re^{i\theta}$ where $(x,y)$ are the rectangular co-ordinates of the complex number and $(r,\theta)$ its polar co-ordinates. We can go from $(x,y)\rightarrow (r,\theta)$ using $r=\sqrt{x^2+y^2}$ and $\theta=\tan^{-1}\frac{y}{x}$ and in the reverse direction $(r,\theta)\rightarrow (x,y)$ using $x=r\cos\theta,\; y =r\sin\theta$. Thus the complex number can be written as $z=r(\cos\theta+i\sin\theta)$. If $r=1$, $z=\cos\theta+i\sin\theta=\boldsymbol{e^{i\theta}}$, given without proof. Hence, we have $e^{-i\theta} = \cos\theta-i\sin\theta$. Through adding and subtracting, we can find that $\cos\theta = \frac{e^{i\theta}+e^{-i\theta}}{2}$ and $\sin\theta = \frac{e^{i\theta}-e^{-i\theta}}{2i}$. Upon dividing the two, $\tan\theta = \dfrac{e^{2i\theta}-1}{i(e^{2i\theta}+1)}=\frac{z-1}{i(z+1)}$, where $z=e^{2i\theta}$. And upon multiplying both top and bottom by $i^2=-1$, we get $\boxed{\tan\theta=\dfrac{i(1-z)}{z+1}}$. Likewise $\boxed{\cot\theta = \frac{i(z+1)}{z-1}}$. ]

Problem attempt : In view of the exponential nature of $e^{2i\theta}=z\Rightarrow z^2=e^{4i\theta}$ and beyond, we have the three terms on the left hand side of our equation as $\tan 2\theta=\dfrac{i(1-z^2)}{z^2+1}$, $\tan 4\theta = \dfrac{i(1-z^4)}{z^4+1}$ and $\cot\theta=\dfrac{i(z^8+1)}{z^8-1}$.

However, in light of the above post #11 by @SammyS, I am going to be satisfied if I can prove that $\cot\theta-\tan\theta = 2\cot 2\theta$. From the results above, we have $$\boldsymbol{\cot 2\theta = \dfrac{i(z^2+1)}{z^2-1}}... (1)$$.

L.H.S. = $\cot\theta-\tan\theta = \dfrac{i(z+1)}{z-1}-\dfrac{i(1-z)}{z+1}=\dfrac{i(z+1)^2+i(z-1)^2}{z^2-1}=\dfrac{2i(z^2+1)}{z^2-1}=2\cot 2\theta$, in light of (1) above.