Exact value of tan10?

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Reply 20

Jtking3000

Original post by Mr M

Do you know about the roots of cubic equations?

Nope

Reply 21

shamika

Original post by Mr M

Nope. I just used my knowledge of obscure trig identities (I like regular polygons).

\alpha=\tan^2 10

\beta=\tan^2 50

\gamma=\tan^2 70

Play around with

\sum \alpha

\sum \alpha \beta

and

\alpha \beta \gamma

I think it would be good if you derive it in full so that people can see what you did, especially as you used something obscure (I would but I don't know what you did either :tongue:

)

Reply 22

Mr M

Study Forum Helper

Original post by shamika

I think it would be good if you derive it in full so that people can see what you did, especially as you used something obscure (I would but I don't know what you did either :tongue:

)

It's really not that interesting.

Sums and products of the squares of trigonometric functions are frequently rational. You can split the sequences into "odd" and "even" multiples.

I knew

\tan^2 10 + \tan^2 30 + \tan^2 50 + \tan^2 70

would be rational but

\tan^2 30

is rational so

\tan^2 10 + \tan^2 50 + \tan^2 70

is rational.

\alpha = \tan^2 10

\beta =\tan^2 50

\gamma = \tan^2 70

\alpha + \beta + \gamma = 9

\alpha \beta + \alpha \gamma + \beta \gamma = 11

\alpha \beta \gamma = \frac{1}{3}

So a cubic with

\tan^2 10

as a root is:

u^3 - 9u^2 + 11u - \frac{1}{3}=0

Multiplying by 3 and substituting

u=x^2

to get

x=\tan 10

as a root:

3x^6 - 27x^4 + 33x^2 - 1=0

Reply 23

marmeduke

Original post by Mr M

Of the things which you say you know are they guesses or do you know them, if you know them, how do you know these things, Ive tried messing around with the sines and coses using formulas i know but cant get to any of the things you give. did you just use a calculator

Reply 24

Perpetuallity

There is no 'exact' value you can find, as

tan(10)

is irrational. However, you can approximate using the taylor expansion.

Reply 25

spex

Original post by Benniboi1

How do you go about finding the exact value of tan10?

I know the 2 triangles for values such as sin45, sin 30, sin60 and the ones for cosine, but how do you find exact values for tan?

Thanks!

IF you mess around with

[br]tan(a+b) = \dfrac{\tan a + \tan b}{1-\tan a \tan b}[br][br]

A useable expression should eventually drop out.

hint

Spoiler

Reply 26

Mr M

Study Forum Helper

Original post by marmeduke

I think I may be confusing you so here is an easier way to derive the same equation.

I expect you know how to obtain this result:

\tan 3x = \frac{3\tan x-\tan^3x}{1-3 \tan^2x}

t = \tan 10

\tan 30 = \frac{3t-t^3}{1-3t^2} = \frac{\sqrt{3}}{3}

Square both sides and the result falls out.

Reply 27

marmeduke

Original post by Mr M

I think I may be confusing you so here is an easier way to derive the same equation.

I expect you know how to obtain this result:

\tan 3x = \frac{3\tan x-\tan^3x}{1-3 \tan^2x}

t = \tan 10

\tan 30 = \frac{3t-t^3}{1-3t^2} = \frac{\sqrt{3}}{3}

Square both sides and the result falls out.

Yeah I get that way, I just want to understand your method. I dont see how to prove the sums you said equal what you say they do

(edited 12 years ago)

Reply 28

Mr M

Study Forum Helper

Original post by marmeduke

Yeah I get that way, I just want to understand your method. I dont see how to prove the sums you said equal what you say they do

Ok they can all be proved but I didn't prove them, I just used intuition and a calculator.

If I wanted to show tan 10 tan 50 tan 70 = tan 30 I would write:

t= tan 10

tan 50 = tan (60 - 10) and express it in terms of t

tan 70 = tan (60 + 10) and express it in terms of t

You will still need to recognise tan 30 in terms of t of course.

Reply 29

marmeduke

Original post by Mr M

O cool, I was hoping for something geometric really you see after you mentioned regular polygons, although thats still quite cool. I just didn't make some sort of obvious connections in my attempts :smile: