Complex Polynomials (1 Viewer)

[Lukybear]

In this question
Deduce
cos4a = 8(cosa)^4 - 8(cosa)^2 + 1
hence solve
8x^4 - 8x^2 +1 = 0

Ive done all except this part:
Deduce exact values of cos(pi/8) and cos (5pi/8)

Im using sum and product of roots and im getting

cos^2(5pi/8)cos^2(pi/8) = 1 and cos^2(5pi/8) + cos^2(pi/8) = 1

But i cannot solve simultaneously as quadratic resulting will give me imaginary.. Why cant i do that?

 

[shaon0]

In this question
Deduce
cos4a = 8(cosa)^4 - 8(cosa)^2 + 1
hence solve
8x^4 - 8x^2 +1 = 0

Ive done all except this part:
Deduce exact values of cos(pi/8) and cos (5pi/8)

Im using sum and product of roots and im getting

cos^2(5pi/8)cos^2(pi/8) = 1 and cos^2(5pi/8) + cos^2(pi/8) = 1

But i cannot solve simultaneously as quadratic resulting will give me imaginary.. Why cant i do that?

Let x=cos@:
8x^4-8x^2+1=0
8(x^2-1/2)^2-1=0
x^2=(1/2)sqrt(2-sqrt(2)) [As cos(5pi/8,pi/8)<1]
Since, cos(5pi/8)<0 and cos(pi/8)>0:
cos(5pi/8)=-(1/2)sqrt(2-sqrt(2)), cos(pi/8)=(1/2)sqrt(2-sqrt(2))
Your equations were wrong, unless i'm wrong.

 

[seanieg89]

 

[shaon0]

Yeah sorry my bad, i did assume they were double roots. I didn't actually read the question fully, just went to solve the poly part.