Seems good...

i hope David McNaughton is well known.

http://dlmcn.com/sin10-irr.html

see dat.

its not that i cant give the proof , but th e article he has given is really good to look at

sin30 = sin3X10

does that solve ur Q ?

wow celes is smarter than DAVID.............hehe

that's where the rational roots theorem comes in handy.

The equation is 8t^3 -6t^2+1 = 0 as in the article linked by b555.

To make it easier consider the reciprocal equation

t^3 -6t^2+8 = 0

By RRT, the only rational roots possible are \pm1, \pm 2, \pm 4, \pm 8 all of which are easy to verify as not being solutions

edit: -6t not -6t²

yes sir i left it as wrk for soumik

i was sure he wud learn more by doing that himself ( cos ive seen some of his gud posts )

we have

8x³ - 6x + 1 = 0

take for contradiction x=p/q

with p , q comprime

we find that cases

p even ,q odd p,q both odd dont satisfy

then in the case p odd , q even we can find that q =2m where m is necessarily odd

again we'll see its odd - odd+odd = 0 which is impossible

wow the rational root theorem is a surprisingly beautiful result

anyone trying to prove it ?

In algebra, the rational root theorem (or 'rational root test') states a constraint on rational solutions (or roots) of the polynomial equation

P(x) = a_nXⁿ + ..... + a₀

with integer coefficients.

If a0 and an are nonzero, then each rational solution x, when written as a fraction x = p/q in lowest terms (i.e., gcd(p,q) = 1) satisfies

p is an integer factor of the constant term a0, and
q is an integer factor of the leading coefficient an.
Thus, a list of possible rational roots of the equation can be derived using the formula .