Sirius Black said:
Martin-are you sure it's the right question coz i substituted roots back to the eqn, it didn't give me zero in the end.
Yeah it looks right to me,
2cos(pi/5) = 1.618...
then 1.618...^4 - 3*1.618...^2 + 1 = 0 (it might come up as something * 10^-5 but that's just rounding errors).
The way I did it in the end was that used in another thread on this board.
Using De Moivres theorem
cos5t + isin5t = (cost + isint)^5
Expand RHS using binomial theorem, then equate real parts, use (sint)^2 = 1 - (cost)^2 to get cos5t as a polynomial in cost.
then can find solutions of cos5t = 1 and cos5t = -1 which gives you surd expressions for cos(Pi/5), cos(2Pi/5), ...
Then you can compare the solutions to z^4-3z^2+1=0 to these expressions and see that they are the same.
So as I said its pretty long and not all that illuminating. I like what Slide Rule said about w+1/w = 2cos(argw) but I don't quite see how to do it.