Polynomials (1 Viewer)

[spice girl]

Originally posted by Dumbarse
so if u want the exact value of cos(pi/32)

put x= pi/128

Don't be stupid! You can put x = pi / 2*4^n ad infinitum...and still won't get an answer...

Now******
cos(4x) = 8cos^4(x) - 8cos^2(x) + 1
Solve the equation 8y^4 - 8y^2 + 1 = 0 and deduce the exact values of cos(pi/8) and cos(5pi/8).
***********

z = cis(x), z^4 = cis(4x)
z^4 = (cosx + isinx)^4
= cos^4(x) + 4isin(x)cos^3(x) - 6sin^2(x)cos^2(x) -4isin^3(x)cos(x) + sin^4(x)
Re(z^4) = cos^4(x) - 6sin^2(x)cos^2(x) + sin^4(x)
= cos^4(x) - 6(1 - cos^2(x))cos^2(x) + (1 - cos^2(x))^2
= 8cos^4(x) - 8cos^2(x) + 1

Consider p(y) = 8y^4 - 8y^2 + 1
Suppose p(y) = 0 has solutions of the form y = cos(x)
Thus p(y) = cos(4x)
Since cos(pi/2) = 0, cos(3pi/2) = 0 then setting 4x = pi/2 then
y = cos(pi/8), cos(3pi/8) are solutions
since p(y) is even, we deduce y = -cos(pi/8), -cos(3pi/8) are also solutions
Now we solve p(y) = 0 using quadratic formula:
y^2 = ( 8 +/- sqrt(64-32) ) / 16
= (2 +/- sqrt(2)) / 4
Thus y = +/-sqrt(2 +/- sqrt(2))/ 2
We know that cos(5pi/8) = -cos(-3pi/8) = -cos(3pi/8)
and that cos(pi/8) > cos(3pi/8) > 0
so we say:
cos(pi/8) = sqrt(2 + sqrt(2)) / 2
cos(5pi/8) = -sqrt(2 - sqrt(2)) / 2

 

[Dumbarse]

yeh, sounds right, my way works also but i couldnt be bothered explaining it

 

[spice girl]

Originally posted by Dumbarse
yeh, sounds right, my way works also but i couldnt be bothered explaining it

yeh right...:rofl:

 

[Yellow]

:jaw:

sorry to be a pain but can you explain:

Since cos(pi/2) = 0, cos(3pi/2) = 0 then setting 4x = pi/2 then
y = cos(pi/8), cos(3pi/8) are solutions
since p(y) is even, we deduce y = -cos(pi/8), -cos(3pi/8) are also solutions
Now we solve p(y) = 0 using quadratic formula:
y^2 = ( 8 +/- sqrt(64-32) ) / 16
= (2 +/- sqrt(2)) / 4
Thus y = +/-sqrt(2 +/- sqrt(2))/ 2
We know that cos(5pi/8) = -cos(-3pi/8) = -cos(3pi/8)
and that cos(pi/8) > cos(3pi/8) > 0
so we say:
cos(pi/8) = sqrt(2 + sqrt(2)) / 2
cos(5pi/8) = -sqrt(2 - sqrt(2)) / 2

a little better. that's where i got lost!

 

[Lazarus]

lol that's like his whole solution =P

 

[spice girl]

Originally posted by Yellow
:jaw:
sorry to be a pain but can you explain:

The stuff you quoted was the mathematics of doing this:

We've p(y) = 8y^4 - 8y^2 + 1
and we've determined that when y = cos(x), p(y) = cos(4x)

When p(y) = 0, cos(4x) = 0

Then ask yourself, for what values of x are there (0 <= x <= pi would suffice) such that cos(4x) = 0 ?

these are: x = pi/8, 3pi/8, 5pi/8, 7pi/8

So cos(pi/8), cos(3pi/8), cos(5pi/8), cos(7pi/8) are zeros of p(y) = 0

Now solve p(y) = 0 using quadratic formula (taking y^2 as the variable for the quadratic)

Since cos(pi/8), cos(3pi/8), cos(5pi/8), cos(7pi/8) are all unique, these must be the only 4 solutions.

Since cos(pi/8) > cos(3pi/8) > cos(5pi/8) > cos(7pi/8), you can get the values you'll find solving the quadratic formula matching up the values of cos(pi/8) and cos(5pi/8)

 

[Dumbarse]

spice girl -"We know that cos(5pi/8) = -cos(-3pi/8) = -cos(3pi/8) "

how>>?

 

[spice girl]

Originally posted by Dumbarse
spice girl -"We know that cos(5pi/8) = -cos(-3pi/8) = -cos(3pi/8) "

how>>?

From the property that cos(x) = -cos(x + pi)
(draw a circle diagram and see for yourself)
Thus, cos(5pi/8) = -cos(-3pi/8)

since cos is an even function
-cos(-3pi/8) = -cos(3pi/8)

 

[Dumbarse]

otay

 

[Yellow]

ohhh now i gets. thanks spice girl.