exact value of tan10 (1 Viewer)

[untouchablecuz]

i was trying to find the exact value of and i ended up with this equation with as a root



any ideas on how to go about solving it?

edit: dw i got it

where

now use the cubic formula

abit tedious, anything else?

 

[khorne]

Yeah, this seems to be the most straightforward way...

As above, 45-30 = 15

Additionally tan(45-30) =/= tan45 - tan 30, but tan45-tan30/1+tan45tan30

 

[untouchablecuz]

tan(10) not tan(15)

and why i need it; i dont really need it, just thought of finding it

 

[khorne]

tan(10) =

(from mathematica)

And those are the only forms...as a root of that polynomial or that

 

[Affinity]

tan(10) =

(from mathematica)

And those are the only forms...as a root of that polynomial or that

That's useless... it's essentially saying tan(10) = sin(10)/cos(10)

 

[untouchablecuz]

LOL that tells me nothing

it simplifies down to tan(10) in like 2 lines

 

[Iruka]

This is quite an interesting question - it is related to the obsession the ancient Greeks had with trisecting the angle using only compass and straight edge techniques.

However, I think you are going to have to use the cubic formula, because as far as I can see, tan(pi/18) is not an element of the field of constructible numbers. If it was, you would be able to express it using the four operations of elementary arithmetic + square roots over the rationals.

 

[untouchablecuz]

this is hopeless, in using the the formula i have to find the cube roots of a complex number and i cant use de'moives because it'll involve trig ratios which i ALSO have to find the exact value of




anyone?

 

[The Nomad]

Cubic formula? Not allowed in HSC...right?

 

[tommykins]

 

[gurmies]

Cubic formula? Not allowed in HSC...right?

Not so much not allowed, as it is not recommended. Any cubic equation you are faced with, will have a lead-in which will ensure an easier path than using Cardano.

 

[MC Squidge]

whats cardano

 

[addikaye03]

All values are in degrees:

tan80=tan(60+20)

=(tan60+tan20)/(1-tan60tan20)

=[√3+(2tan10/1-tan^2(10)]/(1-√3tan20)

Simplify gives: [√3tan^2(10)-2tan(10)-√3]/[tan^2(10)+2√3tan(10)-1]

Since tan80=1/tan10 therefore

tan10=[tan^2(10)+2√3tan(10)-1]/[√3tan^2(10)-2tan(10)-√3]

Alternatively,

I good identity to know it:

tan3x=tanx(tan60-x)(tan60+x)

So let x=10/3

tan10=tan(10/3)(tan60-10/3)(tan60+10/3)

Not that helps alot with this angle lol its good for Q that are like prove: tan50+tan60+tan70=tan80.

It can be easily derived

<SPOILER>

Spoiler

sin 2x = 2. sin x. cos x
So sin 3x = sin (2x + x) = 2s.c^2 + (1-2s^2)s
=2s.(1-s^2) + s - 2s^2
=3s - 4s^3
=s(3 - 4s^2)
=s(3c^2 - s^2).

Sin (60 - x) . sin (60 + x) = 3(c^2 - s^2)/4
=(3-4s^2)/4.

So sin 3x = 4sin x. sin(60-x).sin(60+x).

Same for cosx then tanx=sinx/cosx

<SPOILER>

 

[untouchablecuz]

interesting, thanks

...

eat son

eat