您的位置: 首页  >  作业答案  >  其他  >  在三角形ABC中,a+c=2b,∠a-∠c=60,求sinB的值?sinA + sinC = 2sinB2sin[(A+C)/2] * cos[(A-C)/2] = 2 * 2 * sin(B/2) * cos(B/2)sin[(A+C)/2] * cos30 = 2 * sin(B/2) * cos(B/2)cos[90 - (A+C)/2] * cos30 = 2 * sin(B/2) * cos(B/2)cos(B/2) * cos30 = 2 * sin(B/2) * cos(B/2)cos30 = 2 * sin(B/2)所以,sin(B/2) = 1/2 *cos30 = 根3 / 4为什么cos(B/2)=根13/4

在三角形ABC中,a+c=2b,∠a-∠c=60,求sinB的值?sinA + sinC = 2sinB2sin[(A+C)/2] * cos[(A-C)/2] = 2 * 2 * sin(B/2) * cos(B/2)sin[(A+C)/2] * cos30 = 2 * sin(B/2) * cos(B/2)cos[90 - (A+C)/2] * cos30 = 2 * sin(B/2) * cos(B/2)cos(B/2) * cos30 = 2 * sin(B/2) * cos(B/2)cos30 = 2 * sin(B/2)所以,sin(B/2) = 1/2 *cos30 = 根3 / 4为什么cos(B/2)=根13/4

分类: 作业答案 • 2021-12-18 16:35:47
问题描述：

在三角形ABC中,a+c=2b,∠a-∠c=60,求sinB的值?
sinA + sinC = 2sinB
2sin[(A+C)/2] * cos[(A-C)/2] = 2 * 2 * sin(B/2) * cos(B/2)
sin[(A+C)/2] * cos30 = 2 * sin(B/2) * cos(B/2)
cos[90 - (A+C)/2] * cos30 = 2 * sin(B/2) * cos(B/2)
cos(B/2) * cos30 = 2 * sin(B/2) * cos(B/2)
cos30 = 2 * sin(B/2)
所以,
sin(B/2) = 1/2 *cos30 = 根3 / 4
为什么
cos(B/2)=根13/4

答

因为sinB=2sin(B/2)cos(B/2)=根39/16

答

一个角的正余弦的平方和等于1

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