Question

已知tanx=2,求sin^2x-3sinxcosx+1

Answer 1

我不用万能公式,用一个常规方法吧.
tanx=-2,sinx=-2cosx
X=sin^2x-3sinxcosx+1
=4(cosx)^2+6(cosx)^2+1
X=10(cosx)^2+1(*)
(sinx)^2=4(cosx)^2
2X=20(cosx)^2-6=5(sinx)^2+2
4X=10(sinx)^2+4(**)
(*)+(**)
5x=10+5,X=3
原式=3

Answer 2

tanx=2

sin^2x-3sinxcosx+1
=(1-cos2x)/2-3/2sin2x+1
=1/2-cos2x/2-3/2sin2x+1
=3/2-cos2x/2-3/2sin2x
=3/2-1/2(cos2x+3sin2x)
=3/2-1/2[(1-tan^2x)/(1+tan^2x)+3*2tanx/(1+tan^2x)]
=3/2-1/2[(1-tan^2x+6tanx)/(1+tan^2x)]
=3/2-1/2[(1-4+12)/(1+4)]
=3/2-1/2*9/5
=3/2-9/10
=15/10-9/10
=6/10
=3/5

tanx=2
sinx/cosx=2
sinx=2cosx

sin^2x-3sinxcosx+1
=4(cosx)^2-6(cosx)^2+1
=1-2(cosx)^2
=-cos2x
=-[1-(tanx)^2]/[1+(tanx)^2]
=-(1-4)/(1+4)
=-(-3/5)
=3/5

Answer 3

sinxsinx-3sinxcosx
=(sinxsinx-3sinxcosx)/(sinxsinx+cosxcosx)
分子分母同时除以cosxcosx,得
原式=(tanxtanx-3tanx)/(tanxtanx+1)
=(4-6)/(4+1)=-2/5
所以所求式子值为-2/5+1=3/5