Solving Complex No. Query with Geometric Series and Double Angle Formulae

[sachi]

the first two parts of this question are easy to prove:

1 + cosy/2+ cos(2y)/4 + cos(3y)/8 ... =(4-2cosy)/(5-4cosy)

this can be done using infinite geometric series, and taking the real part.

Then cos(4y) =8({(cosy)^4} -{(cosy)^2}) + 1

which can be done using double angle formulae.

Now we need to solve

x^4 - x^2 + 1/16 = 0

This can be done easily using complex no's, but I'm not sure how to do it using the previous results. I've tried setting cos(4y) = 1/16, and then letting x=cosy, but I end up with some horrible expression involving cos(0.25*arcos(1/16)) which is obviously not what I'm supposed to get.

Thanks

 

[CarlB]

Try putting $$\cos(4y) = 1/2$$.

Carl

 

[Hurkyl]

You notice that:

(cos y)^4 - (cos y)^2 + (1/8)(1 - cos 4y) = 0?

 

[sachi]

thanks v. much. i was probably suffering some sort of miniature brain death by putting cosy = 1/16 (instead of cosy = 1/2!) The trig solution and the completing the square solutions match up as well, so it all work out.