求余弦之和 cos(π／13)+cos(3π／13)+cos(9π／13)

wintex

請問數學

Treenewbee

令\[c_{k} = \cos \frac{k \pi}{13} \]
则有：
\[c_{2k} = 2 c_{k}^2 - 1,   c_{13+k} = c_{13-k}, c_k=c_{26-k}, 2c_{p} c_{q} = c_{p+q} +c_{p-q}\]
易知 \[\sum_{k=1}^6cos\frac{2k \pi}{13}=\frac{\sum_{k=1}^6 2cos\frac{2k \pi}{13}sin\frac{\pi}{13}}{2sin\frac{\pi}{13}}=-\frac{1}{2}\]
即\[c_2+c_4+c_6+c_8+c_{10}+c_{12}=-\frac{1}{2}\]
令\[t=c_1+c_3+c_9=-(c_4+c_{10}+c_{12})\]
显然\[c_2+c_6+c_8=t-\frac{1}{2}\]
\[t^2=((c_4+c_{10}+c_{12}))^2=c_4^2+c_{10}^2+c_{12}^2+2(c_4c_{10}+c_4c_{12}+c_{10}c_{12})\]
\[=\frac{c_8+1}{2}+\frac{c_{20}+1}{2}+\frac{c_{24}+1}{2}+c_{14}+c_6+c_{16}+c_8+c_{22}+c{2}\]
\[=\frac{3}{2}+\frac{3}{2}(c_8+c_6+c_2)+(c_{12}+c_{10}+c_4)\]
\[=\frac{3}{2}+\frac{3}{2}(t-\frac{1}{2})-t\]
\[t^2-\frac{1}{2}t-\frac{3}{4}=0\]
解得：\[t=\frac{1+\sqrt{13}}{4}\]

luyuanhong