spice girl
magic mirror
- Joined
- Aug 10, 2002
- Messages
- 785
Don't be stupid! You can put x = pi / 2*4^n ad infinitum...and still won't get an answer...Originally posted by Dumbarse
so if u want the exact value of cos(pi/32)
put x= pi/128
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).
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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