在三角形ABC中,求证:sin^A/2+sin^B/2+sin^C/2=1-2sinA/2sinB/2sinC/2

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/01 05:26:35
在三角形ABC中,求证:sin^A/2+sin^B/2+sin^C/2=1-2sinA/2sinB/2sinC/2

在三角形ABC中,求证:sin^A/2+sin^B/2+sin^C/2=1-2sinA/2sinB/2sinC/2
在三角形ABC中,求证:sin^A/2+sin^B/2+sin^C/2=1-2sinA/2sinB/2sinC/2

在三角形ABC中,求证:sin^A/2+sin^B/2+sin^C/2=1-2sinA/2sinB/2sinC/2
你的表述出现了一些问题,我想应该是求证:
[sin(A/2)]^2+[sin(B/2)]^2+[sin(C/2)]^2=1-2sin(A/2)sin(B/2)sin(C/2)
若是这样,则方法如下:
在三角形中,有恒等式:cosA+cosB+cosC=1+4sin(A/2)sin(B/2)sin(C/2),
∴1-2[sin(A/2)]^2+1-2[sin(B/2)]^2+1-2[sin(C/2)]^2
 =1+4sin(A/2)sin(B/2)sin(C/2),
∴-2[sin(A/2)]^2-2[sin(B/2)]^2-2[sin(C/2)]^2
 =-2+4sin(A/2)sin(B/2)sin(C/2),
∴[sin(A/2)]^2+[sin(B/2)]^2+[sin(C/2)]^2=1-2sin(A/2)sin(B/2)sin(C/2)
证明完毕.
下面给出恒等式:cosA+cosB+cosC=1+4sin(A/2)sin(B/2)sin(C/2)的证明.
在△ABC中,显然有:
cos[(A+B)/2]=cos[(180°-C)/2]=cos(90°-C/2)=sin(C/2).
∴cosA+cosB+cosC
=2cos[(A+B)/2]cos[(A-B)/2]+1-2[sin(C/2)]^2
=1+2sin(C/2)cos[(A-B)/2]-2sin(C/2)cos[(A+B)/2]
=1+2sin(C/2){cos[(A-B)/2]-cos[(A+B)/2]}
=1+4sin(A/2)sin(B/2)sin(C/2)
注:①若需要求证的实际结论不是这样,请你补充说明.
  ②当出现半角的三角函数时,请不要写成象sinA/2的形式,因为这样容易误解成(1/2)sinA.

sin^A/2 看不懂
还有后面的相乘的式子能写的没有歧义吗?

2sinA/2sinB/2 = cos(A-B)/2 - cos(A+B)/2 = cos(A-B)/2 - sinC/2
右 = 1 - 2sinA/2sinB/2sinC/2 = 1 - ( cos(A-B)/2 - sinC/2 ) sinC/2
= 1 - cos(A-B)/2 * cos(A+B)/2 + (sinC/2)^2
= ...

全部展开

2sinA/2sinB/2 = cos(A-B)/2 - cos(A+B)/2 = cos(A-B)/2 - sinC/2
右 = 1 - 2sinA/2sinB/2sinC/2 = 1 - ( cos(A-B)/2 - sinC/2 ) sinC/2
= 1 - cos(A-B)/2 * cos(A+B)/2 + (sinC/2)^2
= 1 - (1/2) (cosA+cosB) + (1-cosC)/2
= sin^A/2+sin^B/2+sin^C/2 = 左边
注:在三角形ABC中, (A+B+C)/2 = 90° =》cos(A+B)/2 = sinC/2

收起