Using Euler's Formula:
 + isin(\theta))
 + isin(5\theta) = (e^{i\theta})^5 = (cos(\theta) + isin(\theta))^5 )
Expand
 + isin(\theta))^5)
and equate real and imaginary coefficients to
 + isin(5\theta))
Let
 = x)
We know that:
 = 16x^5 - 20x^3 + 5x)
 = 32x^5 - 40x^3 + 10x)
This occurs when
 = 1)
in order to get the -1 that is in the question's polynomial. Solving for this makes:
Therefore:
)
We know that
Therefore, the roots of this equation is going to be x = cos(any angle above) (I think)
 = cos(\pi/3) = 1/2)
1st quadrant. cos(x) > 0
 = cos(7\pi/3) = 1/2)
1st quadrant. cos(x) > 0
 = cos(13\pi/3) = 1/2)
1st quadrant. cos(x) > 0
 = cos(19\pi/3) = 1/2)
1st quadrant. cos(x) > 0
All the above cosine values are equal since the theta value is in the same quadrant and has the same base angle. They all satisfy the need to be equal to 1/2 in order to have the -1 in the polynomial.
We know that
 = 1/2, making \theta = \pi/15)
for the base angle. Therefore, it can be shown that one of the roots of the equation is equal to 1/2 (since cos(pi/3) is one of the roots).
Since all the
)
values are the same, it can be said that:
)
*cos(7\pi/15)*cos(13\pi/15)*cos(19\pi/15) = x*x*-x*-x = (1/2)^4 = 1/16 )
Honestly, probably my best attempt at it, I found it kinda tricky. Good question though. Hopefully someone can provide a more extensive proof.
Home
Help
Search
Login
