Question

已知sinx=2cosx,求(1)（sinx-4cosx)/(5sinx+2cosx) (2)(sinx)^2+2sinxcosx的值

Answer 1

(1)（sinx-4cosx)/(5sinx+2cosx)
=(2cosx-4cosx)/(10cosx+2cosx)
=-2/12
=-1/6
(2)(sinx)^2+2sinxcosx
=4(cosx)^2+4(cosx)^2
=8(cosx)^2
=8/[1+(tanx)^2]
由sinx=2cosx tanx=2
原式=8/[1+2^2]=8/5

Answer 2

sinx=2cosx
tanx=2
(1)（sinx-4cosx)/(5sinx+2cosx) =(tanx-4)/(5tanx+2)=-2/12=-1/6
(2)(sinx)^2+2sinxcosx=[(sinx)^2+2sinxcosx]/(sin^2x+cos^2x)=(tan^2x+2tanx)/(tan^2x+1)=8/5

Answer 3

(1)
（sinx-4cosx)/(5sinx+2cosx)
=（sinx - 2 sinx ) / (5sinx+sinx)
= -sinx / (6 sinx )
= -1/6

(2)
(sinx)^2+2sinxcosx
= (sinx)^2 + 2 cosx sinx
= (sinx)^2 + (sinx)^2
= 2(sinx)^2

Answer 4

sinx=2cosx 且(sinx)^2+(cosx)^2 =1所以(cosx)^2=1/5
(sinx-4cosx)/(5sinx+2cosx)=(-2cosx)/(12cosx ) = -1/6

(sinx)^2 +2sinxcosx = (2cosx)^2 +2*2cosx cosx= 8(cosx )^2 = 8/5 =1.6