Question

3sinx+cosx=0 求下列各式的值？

Answer 1

3sinx+cosx=0
tanx=-1/3
sin²x+cos²x=1
10sin²x=1
sin²x=1/10
1.3cos+5sinx/sinx-cosx
=(3+5tanx)/(tanx-1)
=(3-5/3)/(-1/3-1)
=-1
2 sin²x+2sinxcosx-3cos²x
= sin²x+2sinx(-3sinx)-27 sin²x
=-32sin²x
=-32/10
=-16/5
3:cos2x/(1+sin2x)
=(cos²x-sin²x)/(sinx+cosx)²
=-8sin²x/4sin²x
=-2,9,1.(3cosx+5sinx)/(sinx-cosx)=(-9sinx+5sinx)/(sinx+3sinx)=-1
2. sin²x+2sinxcosx-3cos²x =(sin²x+2sinxcosx-3cos²x )/(sin²x+cos²x )=(sin²x-6sin²x-27sin²x )/...,2,3sinx+cosx=0
3sinx=-cosx
tanx=-1/3
(3) cos2x/(1+sin2x) =[(cosx)^2-(sinx)^2]/[(cosx)^2+(sinx)^2+2sinxcosx]=(cosx-sinx)(cosx+sinx)/(cosx+sinx)^2=(cosx-sinx)/(cosx+sinx)(上下同除cosx)=(1-tanx)/(1...,2,3sinx+cosx=0 求下列各式的值
1：3cos+5sinx/sinx-cosx
2 sin²x+2sinxcosx-3cos²x
3:cos2x/(1+sin2x)