Polynomials (1 Viewer)

[khorne]

Could someone check my reasoning? Sorry for the lack of latex.

Q) Show that cos 4@ = 8 con^4 @ - 8 cos^2 @ + 1, hence:

Solve the equation 8x^4 - 8x^2 +1 = 0, and deduce the value of cos pi/8 and cos 5 pi/8.

Idc about the first part, it's just binomial theorem stuff, but the second part is the working below:

Let x = cos@

therefore cos4@ = pi/2 (My maths teacher said it should be +/- pi/2, but I don't agree, as either way, the angles should be the same.)

therefore:

cos @ = [pi/2 +2 pi k]4 = [pi + 4 pi k]

thus k=0,1,2,3 (4 solutions for quatric):

pi/8, 5pi/8, 9pi/8 and 13pi/8

13pi/8 and 5pi/8 are equal in magnitude, as are pi/8 and 9pi/8

If we solve the equation for x

the answers are:
x = +/- 0.5 sqrt( 2-sqrt(2))
and
x = +/- 0.5 sqrt( 2+sqrt(2))

But, as cos pi/8 > cos 13pi/8 > 0 and 0 > -cos(13pi/8) > cos 9pi/8,

that means cos 13pi/8 must be the small range, i.e x = +/- 0.5 sqrt( 2-sqrt(2)), and as it's equal in magnitude to cos 5pi/8, x = ... is that exact value.

therefore cos pi/8 = x = +/- 0.5 sqrt( 2+sqrt(2))

Comments/suggestions?