7)tan(3π/7)=√7怎样做?
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。
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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。
。
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给出另外一种方法，说实话，这个题目早就超出高考大纲了