$tan^2 10^circ+tan^2 50^circ+tan^2 70^circ=9$ [duplicate]
This question already has an answer here:
Trigonomnetric equality involving tg
1 answer
Strangest thing...*:
$$tan^2 10^circ+tan^2 50^circ+tan^2 70^circ=9tag{1}$$
The trick, as always, is how to prove it.
My idea was to add a "missing" tangent and analyze a similar expression:
$$tan^2 10^circ+tan^2 30^circ+tan^2 50^circ+tan^2 70^circ$$
...and then to attack this sum pairwise (first and the last term, second and third). Despite the fact that I got the same angle ($80^circ$) here and there, I got pretty much nowhere with this approach.
The other interesting fact is that (1) can be rewritten as:
$$cot^2 20^circ+cot^2 40^circ+cot^2 80^circtag{1}$$
...and now the angles are in nice geometric progression. That's the vector of attack that I'm trying to exploit now, but maybe you can entertain youself a little bit too.
*Borrowed from "Usual suspects"
trigonometry
asked Nov 24 at 12:28
Oldboy
6,5101630
marked as duplicate by mathlove, Toby Mak, N. F. Taussig, Jens, Community♦ Nov 24 at 14:18
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
Trigonomnetric equality involving tg
1 answer
Strangest thing...*:
$$tan^2 10^circ+tan^2 50^circ+tan^2 70^circ=9tag{1}$$
The trick, as always, is how to prove it.
My idea was to add a "missing" tangent and analyze a similar expression:
$$tan^2 10^circ+tan^2 30^circ+tan^2 50^circ+tan^2 70^circ$$
...and then to attack this sum pairwise (first and the last term, second and third). Despite the fact that I got the same angle ($80^circ$) here and there, I got pretty much nowhere with this approach.
The other interesting fact is that (1) can be rewritten as:
$$cot^2 20^circ+cot^2 40^circ+cot^2 80^circtag{1}$$
...and now the angles are in nice geometric progression. That's the vector of attack that I'm trying to exploit now, but maybe you can entertain youself a little bit too.
*Borrowed from "Usual suspects"
trigonometry
asked Nov 24 at 12:28
Oldboy
6,5101630
marked as duplicate by mathlove, Toby Mak, N. F. Taussig, Jens, Community♦ Nov 24 at 14:18
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
3
This is a very similar question.
– Toby Mak
Nov 24 at 12:43
add a comment |
This question already has an answer here:
Trigonomnetric equality involving tg
1 answer
Strangest thing...*:
$$tan^2 10^circ+tan^2 50^circ+tan^2 70^circ=9tag{1}$$
The trick, as always, is how to prove it.
My idea was to add a "missing" tangent and analyze a similar expression:
$$tan^2 10^circ+tan^2 30^circ+tan^2 50^circ+tan^2 70^circ$$
...and then to attack this sum pairwise (first and the last term, second and third). Despite the fact that I got the same angle ($80^circ$) here and there, I got pretty much nowhere with this approach.
The other interesting fact is that (1) can be rewritten as:
$$cot^2 20^circ+cot^2 40^circ+cot^2 80^circtag{1}$$
...and now the angles are in nice geometric progression. That's the vector of attack that I'm trying to exploit now, but maybe you can entertain youself a little bit too.
*Borrowed from "Usual suspects"
trigonometry
asked Nov 24 at 12:28
Oldboy
6,5101630
This question already has an answer here:
Trigonomnetric equality involving tg
1 answer
Strangest thing...*:
$$tan^2 10^circ+tan^2 50^circ+tan^2 70^circ=9tag{1}$$
The trick, as always, is how to prove it.
My idea was to add a "missing" tangent and analyze a similar expression:
$$tan^2 10^circ+tan^2 30^circ+tan^2 50^circ+tan^2 70^circ$$
...and then to attack this sum pairwise (first and the last term, second and third). Despite the fact that I got the same angle ($80^circ$) here and there, I got pretty much nowhere with this approach.
The other interesting fact is that (1) can be rewritten as:
$$cot^2 20^circ+cot^2 40^circ+cot^2 80^circtag{1}$$
...and now the angles are in nice geometric progression. That's the vector of attack that I'm trying to exploit now, but maybe you can entertain youself a little bit too.
*Borrowed from "Usual suspects"
This question already has an answer here:
Trigonomnetric equality involving tg
1 answer
trigonometry
trigonometry
asked Nov 24 at 12:28
Oldboy
6,5101630
asked Nov 24 at 12:28
Oldboy
6,5101630
asked Nov 24 at 12:28
Oldboy
6,5101630
asked Nov 24 at 12:28
Oldboy
6,5101630
asked Nov 24 at 12:28
Oldboy
6,5101630
6,5101630
marked as duplicate by mathlove, Toby Mak, N. F. Taussig, Jens, Community♦ Nov 24 at 14:18
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by mathlove, Toby Mak, N. F. Taussig, Jens, Community♦ Nov 24 at 14:18
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
3
This is a very similar question.
– Toby Mak
Nov 24 at 12:43
add a comment |
3
This is a very similar question.
– Toby Mak
Nov 24 at 12:43
3
3
This is a very similar question.
– Toby Mak
Nov 24 at 12:43
This is a very similar question.
– Toby Mak
Nov 24 at 12:43
add a comment |
1 Answer
1
active
oldest
votes
You can write it as
$$cot^220^circ+cot^240^circ+cot^260^circ+cdots+cot^2160^circ=frac{56}3$$
That has eight multiples of $180^circ/9$, and you can find a similar equation for other numbers instead of $9$.
answered Nov 24 at 12:53
Empy2
33.4k12261
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can write it as
$$cot^220^circ+cot^240^circ+cot^260^circ+cdots+cot^2160^circ=frac{56}3$$
That has eight multiples of $180^circ/9$, and you can find a similar equation for other numbers instead of $9$.
answered Nov 24 at 12:53
Empy2
33.4k12261
add a comment |
You can write it as
$$cot^220^circ+cot^240^circ+cot^260^circ+cdots+cot^2160^circ=frac{56}3$$
That has eight multiples of $180^circ/9$, and you can find a similar equation for other numbers instead of $9$.
answered Nov 24 at 12:53
Empy2
33.4k12261
add a comment |
You can write it as
$$cot^220^circ+cot^240^circ+cot^260^circ+cdots+cot^2160^circ=frac{56}3$$
That has eight multiples of $180^circ/9$, and you can find a similar equation for other numbers instead of $9$.
answered Nov 24 at 12:53
Empy2
33.4k12261
You can write it as
$$cot^220^circ+cot^240^circ+cot^260^circ+cdots+cot^2160^circ=frac{56}3$$
That has eight multiples of $180^circ/9$, and you can find a similar equation for other numbers instead of $9$.
answered Nov 24 at 12:53
Empy2
33.4k12261
answered Nov 24 at 12:53
Empy2
33.4k12261
answered Nov 24 at 12:53
Empy2
33.4k12261
answered Nov 24 at 12:53
Empy2
33.4k12261
33.4k12261
add a comment |
add a comment |
3
This is a very similar question.
– Toby Mak
Nov 24 at 12:43