证明：cos(π/7)cos(2π/7)cos(3π/7) =2³sin(π/7)cos(π/7)cos(2π/7)cos(3π/7)/[2³sin(π/7)] =2²sin(2π/7)cos(2π/7)cos(3π/7)/[8sin(π/7)] =2sin(4π/7)cos(3π/7)/[8sin(π/7)] =2sin(3π/7)cos(3π/7)/[8sin(π/7)] =sin(6π/7)/[8sin(π/7)] =1/8 欲证tan(π/7)tan(2π/7)tan(3π/7)=√7 只需8sin(π/7)sin(2π/7)sin(3π/7)=√7 右端是无理数，直接证明有困难，考虑平方， 64sin³(π/7)sin³(2π/7)sin³(3π/7)=7 即8[1-cos(2π/7)][1-cos(4π/7)][1-cos(6π/7)]=7 亦即8[1-(cos(2π/7)+cos(4π/7)+cos(6π/7))+cos(2π/7)cos(4π/7)+cos(4π/7)cos(6π/7)+cos(6π/7)cos(2π/7)-cos(2π/7)cos(4π/7)cos(6π/7)]=7 而cos(2π/7)cos(4π/7)cos(6π/7)=cos(2π/7)cos(3π/7)cos(π/7)=1/8 故只需证 cos(2π/7)cos(4π/7)+cos(4π/7)cos(6π/7)+cos(6π/7)cos(2π/7) =cos(2π/7)+cos(4π/7)+cos(6π/7) 左边=cos(4π/7)[cos(2π/7)+cos(6π/7)]+cos(6π/7)cos(2π/7) =cos(4π/7)2cos(4π/7)cos(2π/7)+cos(6π/7)cos(2π/7) =cos(2π/7)×2cos²(4π/7)+cos(6π/7)cos(2π/7) =cos(2π/7)(1+cos(8π/7))+cos(6π/7)cos(2π/7) =cos(2π/7)+cos(2π/7)[cos(8π/7))+cos(6π/7)] =cos(2π/7)+cos(2π/7)×2cosπcos(π/7) =cos(2π/7)-2cos(2π/7)cos(π/7) =cos(2π/7)-(cos(3π/7)+cos(π/7)) =cos(2π/7)+cos(4π/7)+cos(6π/7)=右边 证毕。
以上是证明过程，如要求解，可能需要通过构造图形来解决，这方面我不太在行。
这种题直接忽视,高考绝对没有!
证明三角恒等式 证明tan(π/7)*tan (2π/7)*tan (3π/7)=√7。 证明 令A=π/7,B=2π/7,C=4π/7,那么A+B+C=π， 设三角形ABC对应边长分别a,b,c。 根据正弦定理:a/sin(π/7)=b/sin(2π/7)=c/sin(4π/7) 即可求得:cos(π/7)=b/(2a), cos(2π/7)=c/(2b), cos(4π/7)=-a/(2c)， 再注意到倍角公式: ∠B=2∠A b^2-a^2=ac，(1) ∠C=2∠B c^2-b^2=ab, (2) (1)+(2)得:c^2-a^2=a(b+c)， 由此可推出:bc=a(b+c), c^2-a^2=bc [sin(π/7)]^2=1-[cos(π/7)]^2=1-b^2/(4a^2)=(4a^2-b^2)/(4a^2) =(3a^2-ac)/(4a^2)=(3a-c)/(4a)。
同理可得: [sin(2π/7)]^2=(3b-a)/(4b), [sin(4π/7)]^2=(3c+b)/(4c)。 所以 (64abc) *[sin(π/7)*sin (2π/7)*sin (4π/7)]^2 =(3a-c)*(3b-a)*(3c+b)=(3a-c)*(9bc-3ac+3b^2-ab) =(3a-c)*(6bc+3b^2+2ab)=b*(3a-c)*(6c+2a+3b) =b*(16ac-6c^2+6a^2+9ab-3bc) =b*(16ac-9bc+9ab)=b*(16ac-9ac)=7abc 故sin(π/7)*sin (2π/7)*sin (3π/7)= (√7)/8 而cos(π/7)*cos(2π/7)*cos(3π/7)=-cos(π/7)*cos(2π/7)*cos(4π/7) =[b/(2a)]*[c/(2b)]*[a/(2c)]=1/8 因此tan(π/7)*tan (2π/7)*tan (3π/7)=√7。
证毕。 。
给出另外一种方法，说实话，这个题目早就超出高考大纲了
答：证明tan(π/7)*tan (2π/7)*tan (3π/7)=√7 一般地，我们有: tan[π/(2n+1)]*tan[2π/(2n+1)]*…*tan[...详情>>
