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用tAnx表示sin2x,Cos2x

根据万能公式得 sin2x=2tanx/(1+tan??x),cos2x=(1-tan??x)/(1+tan??x),

tanx=sinx/cosx=2sinx/2sinxcosx=(1-cos2x)/sin2x 或者=2sinxcosx/2cosx=sin2x/(1+cos2x)

sin2x=2sinxcosx=2sinxcosx/(sinx^2+cosx^2) 上下同除以cosx^2 sin2x=2tanx/(1+tanx^2) cos2x=cosx^2-sinx^2/(cosx^2+sinx^2) 同除以cosx^2 =(1-tanx^2)/(1+tanx^2)

sin2x/cos2x等于tan 2x

解:因为tanx=2所以tan2x=2tanx/[1-(tanx)^2]=2*2/(1-2^2)=-4/3所以(sin2x+cos2x)=[(sin2x/cos2x)+(cos2x)/(cos2x)]=(tan2x+1)=[-4/3+1]=-1/3

证明:cos2x = (cosx)^2-(sinx)^2=2*(cosx)^2 -1 1-cos2x=1-[2*(cosx)^2 -1]=2-2*(cosx)^2=2*[1-(cosx)^2]=2*(sinx)^2sin2x=2cosx*sinx所以:(1-cos2x)/sin2x=[2*(sinx)^2]/2cosx*sinx=sinx/cos=tanx

先化简cos2x/1+sin2x=(1-tanx)/(1+tanx) 再用tanx/2=t求tanx tanx=(2tanx/2)/1-(tanx/2)此处二倍角 tanx=2t/(1-t^2)代入cos2x/1+sin2x=(1-tanx)/(1+tanx) 所以cos2x/1+sin2x=(1-t^2-2t)/(1-t^2+2t) 哪不懂可追问

1+sinx=2cos^2 (丌/4-x/2)1-sinx=2sin^2 (丌/4-x/2)sin2x=2tanx/(1+tan^2 x)cos2x=(1-tan^2 x)/(1+tan^2 x)

用万能公式f(tanx)=sin2x+cos2x=[2tanx+1-tan^2(x)]/[1+tan^2(x)]f(x)=(2x+1-x^2)/(1+x^2)

万能公式sin2x=2tanx/1+(tanx)^2=3/5 cos2x=1-(tanx)^2/1+(tanx)^2=-4/5 所以原式=7/5

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