高一数学三角函数及定理题,回答必须有过程,谢谢.只问第八题!
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高一数学三角函数及定理题,回答必须有过程,谢谢.只问第八题!
![高一数学三角函数及定理题,回答必须有过程,谢谢.只问第八题!](/uploads/image/z/5975956-28-6.jpg?t=%E9%AB%98%E4%B8%80%E6%95%B0%E5%AD%A6%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B0%E5%8F%8A%E5%AE%9A%E7%90%86%E9%A2%98%2C%E5%9B%9E%E7%AD%94%E5%BF%85%E9%A1%BB%E6%9C%89%E8%BF%87%E7%A8%8B%2C%E8%B0%A2%E8%B0%A2.%E5%8F%AA%E9%97%AE%E7%AC%AC%E5%85%AB%E9%A2%98%21)
第8题答案是4
(注:江苏教材一般写cosA,不写cos∠A,其他地区可能略有差异)
由余弦定理,cosC=(a^2+b^2-c^2)/(2ab),
则原式b/a+a/b=6cosC=6(a^2+b^2-c^2)/(2ab),
得3c^2=2a^2+2b^2,(a^2+b^2)=3/2c^2,
由正弦定理,a/sinA=b/sinB=c/sinC=2R,(R为三角形ABC外接圆半径),
则待求的式子用切割化弦可得:
tanC/tanA+tanC/tanB
=(sinC/cosC)*(cosA/sinA+cosB/sinB)
=(sinC/cosC)*[(sinB*cosA+sinA*cosB)/(sinA*sinB)]
=(sinC/cosC)*[sin(A+B)/(sinA*sinB)]
=(sinC/cosC)*[sin(π-C)/(sinA*sinB)]
=(sinC/cosC)*[sinC/(sinA*sinB)]
=[(sinC*sinC)/(sinA*sinB)]/cosC
=[(c/2R)*(c/2R)]*[(a/2R)*(b/2R)]/cosC
=c^2*ab/[(a^2+b^2-c^2)/(2ab)]
=2c^2/(3/2c^2-c^2)
=2c^2/(1/2c^2)
=4
即待求的式子值为4,
(注:江苏教材一般写cosA,不写cos∠A,其他地区可能略有差异)
由余弦定理,cosC=(a^2+b^2-c^2)/(2ab),
则原式b/a+a/b=6cosC=6(a^2+b^2-c^2)/(2ab),
得3c^2=2a^2+2b^2,(a^2+b^2)=3/2c^2,
由正弦定理,a/sinA=b/sinB=c/sinC=2R,(R为三角形ABC外接圆半径),
则待求的式子用切割化弦可得:
tanC/tanA+tanC/tanB
=(sinC/cosC)*(cosA/sinA+cosB/sinB)
=(sinC/cosC)*[(sinB*cosA+sinA*cosB)/(sinA*sinB)]
=(sinC/cosC)*[sin(A+B)/(sinA*sinB)]
=(sinC/cosC)*[sin(π-C)/(sinA*sinB)]
=(sinC/cosC)*[sinC/(sinA*sinB)]
=[(sinC*sinC)/(sinA*sinB)]/cosC
=[(c/2R)*(c/2R)]*[(a/2R)*(b/2R)]/cosC
=c^2*ab/[(a^2+b^2-c^2)/(2ab)]
=2c^2/(3/2c^2-c^2)
=2c^2/(1/2c^2)
=4
即待求的式子值为4,