问一下,用TAN B/2 咋表示COSB SINB TANB 咋办?要有过程
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/15 19:01:04
问一下,用TAN B/2 咋表示COSB SINB TANB 咋办?要有过程
sin²(b/2)+cos²(b/2)=1
所以1/sinb=[sin²(b/2)+cos²(b/2)]/2sin(b/2)cos(b/2)
=sin²(b/2)/[2sin(b/2)cos(b/2)]+cos²(b/2)/[2sin(b/2)cos(b/2)]
=sin(b/2)/2cos(b/2)+cos(b/2)/2cos(b/2)
=tan(b/2)+1/2tan(b/2)
=[2tan²(b/2)+1]/2tan(b/2)
所以sinb=2(b/2)/[1+2tan²(b/2)]
cosb=cos²(b/2)-sin²(b/2)=[cos²(b/2)-sin²(b/2)]/[cos²(b/2)+sin²(b/2)]
上下除以cos²(b/2)
所以cosb=[1-tan²(b/2)]/[1+tan²(b/2)]
tanb直接公式
tanb=2tan(b/2)/[1-tan²(b/2)]
所以1/sinb=[sin²(b/2)+cos²(b/2)]/2sin(b/2)cos(b/2)
=sin²(b/2)/[2sin(b/2)cos(b/2)]+cos²(b/2)/[2sin(b/2)cos(b/2)]
=sin(b/2)/2cos(b/2)+cos(b/2)/2cos(b/2)
=tan(b/2)+1/2tan(b/2)
=[2tan²(b/2)+1]/2tan(b/2)
所以sinb=2(b/2)/[1+2tan²(b/2)]
cosb=cos²(b/2)-sin²(b/2)=[cos²(b/2)-sin²(b/2)]/[cos²(b/2)+sin²(b/2)]
上下除以cos²(b/2)
所以cosb=[1-tan²(b/2)]/[1+tan²(b/2)]
tanb直接公式
tanb=2tan(b/2)/[1-tan²(b/2)]
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