求三角形三角函数证明题做法,
来源:学生作业帮 编辑:搜狗做题网作业帮 分类:数学作业 时间:2024/08/11 15:57:29
求三角形三角函数证明题做法,
证明:(1) 对任意三角形ABC,tanA/2*tanB/2+tanB/2*tanC/2+tanC/2*tanA/2=1 (2) 在非直角三角形ABC,tanA+tanB+tanC=tanA*tanB*tanC
证明:(1) 对任意三角形ABC,tanA/2*tanB/2+tanB/2*tanC/2+tanC/2*tanA/2=1 (2) 在非直角三角形ABC,tanA+tanB+tanC=tanA*tanB*tanC
![求三角形三角函数证明题做法,](/uploads/image/z/16142381-53-1.jpg?t=%E6%B1%82%E4%B8%89%E8%A7%92%E5%BD%A2%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B0%E8%AF%81%E6%98%8E%E9%A2%98%E5%81%9A%E6%B3%95%2C)
(1)tanA/2×tanB/2+tanB/2×tanC/2+tanC/2×tanA/2
=tanA/2×tanB/2+tanC/2×(tanA/2+tanB/2)
=tanA/2×tanB/2+tan[90-(A+B)/2]×(tanA/2+tanB/2)
=tanA/2×tanB/2+cot(A/2+B/2)×(tanA/2+tanB/2)
=tanA/2×tanB/2+(tanA/2+tanB/2)/tan(A/2+B/2)
=tanA/2×tanB/2+1-tanA/2×tanB/2
=1
(2)∵tan(A+B)=tanA+tanB/1-tanA*tanB
tan(A+B)=tan(π-C)=-tanC
∴tanA+tanB/1-tanA*tanB=-tanC
整理移项即得
tanA+tanB+tanC=tanA*tanB*tanC
=tanA/2×tanB/2+tanC/2×(tanA/2+tanB/2)
=tanA/2×tanB/2+tan[90-(A+B)/2]×(tanA/2+tanB/2)
=tanA/2×tanB/2+cot(A/2+B/2)×(tanA/2+tanB/2)
=tanA/2×tanB/2+(tanA/2+tanB/2)/tan(A/2+B/2)
=tanA/2×tanB/2+1-tanA/2×tanB/2
=1
(2)∵tan(A+B)=tanA+tanB/1-tanA*tanB
tan(A+B)=tan(π-C)=-tanC
∴tanA+tanB/1-tanA*tanB=-tanC
整理移项即得
tanA+tanB+tanC=tanA*tanB*tanC