已知tanx/2=根号5,求(1+sinx-cosx)/(1+sinx+cosx)的值!
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/10 23:43:04
已知tanx/2=根号5,求(1+sinx-cosx)/(1+sinx+cosx)的值!
sinx = 2tan(x/2)/(1+tanx/2tanx/2)
= 2*sqrt(5)/(1+5)
= sqrt(5)/3
cosx = (1-tanx/2tanx/2)/(1+tanx/2tanx/2)
= -2/3
(1+sinx-cosx)/(1+sinx+cosx)
= (1+ sqrt(5)/3+2/3)/(1+ sqrt(5)/3-2/3)
= sqrt(5)
注: sqrt(5)为根号5
或者(为表达方便令t=tanx/2)
sina = 2t/(1+t*t)
cosa = (1-t*t)/(1+t*t)
代人有
(1+sinx-cosx)/(1+sinx+cosx) = 2t(t+1)/[2(t+1)]
= t
= tanx/2
= 根号5
= 2*sqrt(5)/(1+5)
= sqrt(5)/3
cosx = (1-tanx/2tanx/2)/(1+tanx/2tanx/2)
= -2/3
(1+sinx-cosx)/(1+sinx+cosx)
= (1+ sqrt(5)/3+2/3)/(1+ sqrt(5)/3-2/3)
= sqrt(5)
注: sqrt(5)为根号5
或者(为表达方便令t=tanx/2)
sina = 2t/(1+t*t)
cosa = (1-t*t)/(1+t*t)
代人有
(1+sinx-cosx)/(1+sinx+cosx) = 2t(t+1)/[2(t+1)]
= t
= tanx/2
= 根号5
已知tanx/2=根号5,求(1+sinx-cosx)/(1+sinx+cosx)的值!
已知sinx+cosx=(根号3+1)/2,求sinx/(1-1/tanx)+cosx/(1-tanx)的值
已知sinx+cosx=-根号10/5,求(1)1/sinx+1/cosx (2)tanx的值
已知cosx=根号1+sinx-1-根号sinx/2,求tanx的值
已知(1-tanx)/(1+tanx)=2,求(sinx-cosx)/(sinx+cosx)的值
已知tanx=根号2,求2cos2(x/2)-sinx-1/sinx+cosx的值
已知sinx+cosx=(根号3-1)/2,求tanx
:已知(sinx+cosx)/(sinx-cosx)=3,求tanx,2sin2x+(sinx-cosx)2的值.
已知tanx=1/2 (1) 求sinx-cosx/2sinx+3cosx的值 (2) 求sinx cosx的值
已知sinx+cosx=1/5,且0<x<π.(1) 求sinx、cosx、tanx的值.
已知(tanx)\(tanx-1)=-1,求下列各式的值:(1)(sinx-3cosx)\(sinx+cosx) (2)
(1)已知:(4sinx-2cosx)/(5cosx+3sinx)=6/11 求sinx乘以cosx的值