This question is killing me (1 Viewer)

[tiggerfamilytre]

This comes from Patel's book:

"Find the roots of z^6 + 1 = 0 and hence resolve z^6 + 1 into real quadratic factros; deduce that cos3x = 4(cosx - cos[pi/6])(cosx - cos[pi/2])(cosx - cos[{5pi}/6])"

Please help

 

[dawso]

z^6 = -1

one root is i, other roots are equally spaced around unit circle, from that get the factors, um, will work on other bit now...

 

[Slidey]

Scrap that.
(z^2-1)(z^4-z^2+1)
(z^2+1)^2-(sqrt3.z)^2=z^4-z^2+1
z^4-z^2+1=(z^2-sqrt3.z+1)(z^2+sqrt3.z+1)
(z^2-1)(z^2-sqrt3.z+1)(z^2+sqrt3.z+1) let divide through by z^3:
z^3+1/z^3=(z-1/z)(z+1/z-sqrt3)(z+1/z+sqrt3)
z+1/z=2cos@ (for z=cos@+isin@)
2cos3@=2cos@(2cos@-sqrt3)(2cos@+sqrt3)
1/4cos3@=cos@(cos@-(sqrt3)/2)(cos@+(sqrt3)/2)
To get the pi/6 and stuff, go cos(pi/6)=sqrt(3)/2, cos(p/2)=0 (as you can see this is a rather involved question)
cos3@=4(cos@-cos(pi/2))(cos@-cos(5pi/6))(cos@-cos(pi/6))

 

[tiggerfamilytre]

wow.

That's awesome. Thank you so much.

 

[Tomahawk1801]

tysm

 

[5uckerberg]

tysm

LMAO.