已知向量a=(sinx,cosx)b=(根号3cosx,cosx)且b不等于0 函数f(x)=2a·b-1 ,若a‖b,求cos2x/[f(x)+1]的值
1个回答
展开全部
if a//b
then
cosx/sinx =cosx/√3cosx
tanx = √3
tan2x = 2√3/(1-3) = -√3
a.b
=(sinx,cosx).(√3cosx,cosx)
= √3sinxcosx + (cosx)^2
f(x)
= 2a.b-1
=2(√3sinxcosx + (cosx)^2)-1
=√3sin2x - cos2x
cos2x/[f(x)+1]
= cos2x/( √3sin2x - cos2x)
= 1/(√3tan2x-1)
=1/(-3-1) = -1/4
then
cosx/sinx =cosx/√3cosx
tanx = √3
tan2x = 2√3/(1-3) = -√3
a.b
=(sinx,cosx).(√3cosx,cosx)
= √3sinxcosx + (cosx)^2
f(x)
= 2a.b-1
=2(√3sinxcosx + (cosx)^2)-1
=√3sin2x - cos2x
cos2x/[f(x)+1]
= cos2x/( √3sin2x - cos2x)
= 1/(√3tan2x-1)
=1/(-3-1) = -1/4
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询