根据正弦定理 a/sinA=b/sinB=c/sinC
得:a=(sinA/sinB)*b c=(sinC/sinB)*b
将其带入已知条件 a+c=2b中
可得sinA+sinC=2sinB
根据三角函数和公式
sinA+sinC=2sin[(A+C)/2] * cos[(A-C)/2]
这一步
sinA+sinC=2sinB
根据三角函数和公式
sinA+sinC=2sin[(A+C)/2] * cos[(A-C)/2]
怎么得出的?请详细解释一下,根据什么公式?这个公式是什么
在三角形ABC中,设a+c=2b,A-C=60度,求sinB的值
答案:2 悬赏:50 手机版
解决时间 2021-04-06 14:24
- 提问者网友:相思似海深
- 2021-04-05 16:43
最佳答案
- 五星知识达人网友:一把行者刀
- 2021-04-05 16:48
sin(A + B) + sin(A - B) = sin(A)cos(B) + cos(A)sin(B) + sin(A)cos(B)
- cos(A)sin(B)
= 2sin(A)cos(B) ……………………………… ①
令A = (a + c)/2, B = (a - c)/2 则:
sin(A + B) + sin(A - B) = sin[(a + c)/2 + (a - c)/2]
+ sin[(a + c)/2 - (a - c)/2]
= sin(a) + sin(c);
2sin(A)cos(B) = 2sin[(a + c)/2] * cos[(a - c)/2]
由①得 sin(a) + sin(c) = 2sin[(a + c)/2] * cos[(a - c)/2]
- cos(A)sin(B)
= 2sin(A)cos(B) ……………………………… ①
令A = (a + c)/2, B = (a - c)/2 则:
sin(A + B) + sin(A - B) = sin[(a + c)/2 + (a - c)/2]
+ sin[(a + c)/2 - (a - c)/2]
= sin(a) + sin(c);
2sin(A)cos(B) = 2sin[(a + c)/2] * cos[(a - c)/2]
由①得 sin(a) + sin(c) = 2sin[(a + c)/2] * cos[(a - c)/2]
全部回答
- 1楼网友:詩光轨車
- 2021-04-05 17:16
a=bsina/sinb c=bsinc/sinb
a+c=(b/sinb)(sina+sinc)=2b
sina+sinc=2sinb
sina+sinc=2sin[(a+c)/2]cos[(a-c)/2]=2sin(90-b/2)cos30=√3cosb/2
2sinb=4sinb/2cosb/2
sinb/2=√3/4 cosb/2=√13/4
sinb=√39/8
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