分类：教育/科学-理工学科-数学
问题：求三角函数值求H=1/[sin(π/7)]^2+1/[sin(2? 
求三角值求H=1/[sin(π/7)]^2+1/[sin (2π/7)]^2+1/[sin (4π/7)]^2的值。
回答：求三角值 求H=1/[sin(π/7)]^2+1/[sin (2π/7)]^2+1/[sin (4π/7)]^2的值。 解  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-ab-ac=0,     (3)[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).H=4[a/(3a-c)+b/(3b-a)+c/(3c+b)] ，注意到:bc=ab+ac，上式化简即得H=1/[sin(π/7)]^2+1/[sin (2π/7)]^2+1/[sin (4π/7)]^2=8. 证毕.