Question

高一三角函数转换问题

我看到了这公式.可不知道怎么证明.好想知道过程请大家证明的详细点.在三角形ABC中SinA+sinB+sinC=4cos(A/2)cos(B/2)cos(C/2).还有一个也是在三角形中的tanAtanBtanC=tanA+tanB+tanC
虽然你后面还有点小错。不过我还是看懂了，谢了

Answer 1

SinA+sinB+sinC
=2sin(A+B)/2cos(A-B)/2+2sin(C/2)cos(C/2)
=2cos(C/2)cos(A-B)/2+2cos(A+B)/2cos(C/2)
=2cos(C/2)[cos(A-B)/2+cos(A+B)/2]
=2cos(C/2)cos(A/2)cos(B/2)
得证
tanC=-tan(A+B)=-(tanA+tanB)/(1-tanAtanB)
去分母
tanC-tanAtanBtanC=-tanA-tanB
所以tanAtanBtanC=tanA+tanB+tanC
得证
希望能帮到你，祝学习进步O(∩_∩)O

Answer 2

sinA+sinB+sinC=sinA+sinB+sin(π-A-B)=sinA+sinB+sin(A+B)=sinA+sinB+sinAcosB+cosAsinB
=sinA(1+cosB)+sinB(1+cosA)=2sin(A/2)*cos(A/2)*2cos²(A/2)+2sin(B/2)*cos(B/2)*2cos²(A/2)
=4cos(A/2)*cos(B/2)[sin(A/2)cos(A/2)+cos(A/2)sin(A/2)]=4cos(A/2)*cos(B/2)*sin[(A+B)/2]
=4cos(A/2)*cos(B/2)*sin(π/2-C/2)=4cos(A/2)*cos(B/2)*sin(C/2)。
tanA+tanB+tanC=tanA+tanB+tan(π-A-B)=tanA+tanB-tan(A+B)
=tanA+tanB-[tan(A)+tan(B)]/(1-tanAtanB)=-tanAtanB(tanA+tanB)/(1-tanAtanB)
=-tanAtanBtan(A+B)=tanAtanBtanC