Simplify ∏k=15tan⁡kπ11 and ∑k=15tan2kπ11

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Larry Hogan · Accepted Answer

The main building block of our solution will be the formula∏k=0N−1(x−e2ktπN)=xN−1 (0)It will be convenient to rewrite (0) for odd N=2n+1∏k=1n[x2+1−2xcos⁡πk2n+1]=x2n+1−1x−1 (1)Replacing therein x→−x and multiplying the result by (1), we may also write∏k=1n[(x2−1)2+4x2sin2πk2n+1]=1−x4n+21−x2 (2)1. Setting in (1) x=−i, we get(2i)n∏k=1ncos⁡πk2n+1=i2n+1−1i−1⇒∏k=1n2cos⁡πk2n+1=12. Setting in (2) x=1 and computing the corresponding limit on the right, we get∏k=1n2sin⁡πk2n+1=[limx→11−x4n+21−x2]12=2n+13. Combining the two results yields∏k=1ntan⁡πk2n+1=2n+1and to find X, it suffices to set n=5.4. To find Y, let us rewrite (1) in the form (set x=−eiy)∏k=1n[2cos⁡γ+2cos⁡πk2n+1]=cos⁡(2n+1)γ2cos⁡γ2Taking the logarithm and differentiating twice with respect to γ, we find∑k=1n1(cos⁡γ+cos⁡πk2n+1)2=−1sin⁡γddy(1sin⁡γddyln⁡cos⁡(2n+1)γ2cos⁡γ2) (3)5. Computing the right side of (3) and setting therein γ=π2, we finally arrive at⇒∑k=1ntan2πk2n+1=n(2n+1)This yields Y=55

Simplify \prod_{k=1}^5\tan\frac{k\pi}{11} and \sum_{k=1}^5\tan^2\frac{k\pi}{11}

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