tan(π+x)+cos(3π+2)=2,则sinxcosx的值等于

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tan(π+x)+cos(3π+2)=2,则sinxcosx的值等于
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tan(π+x)+cos(3π+2)=2,则sinxcosx的值等于
tan(π+x)+cos(3π+2)=2,则sinxcosx的值等于

tan(π+x)+cos(3π+2)=2,则sinxcosx的值等于
tan(π+x)+cos(3π+2)=2
tanx-cos2=2
tanx=2+cos2
1+(tanx)^2=1/(cosx)^2
sinxcosx
=sinx/cosx*(cosx)^2
=tanx*1/[1+(tanx)^2]
=(2+cos2)/[1+(2+cos2)^2]
=(2+cos2)/[5+4cos2+(cos2)^2]

tan(π+x)+cos(3π+2)
tanx-cos2=2
tanx=2+cos2
sinx=(2+cos2)cosx
sin^2x=(2+cos2)^2cos^2x
1-cos^2=(2+cos2)^2cosx^2
cos^2x=1/[(2+cos2)^2+1]
cosx=√1/[(2+cos2)^2+1]
sin^2x=(2+c...

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tan(π+x)+cos(3π+2)
tanx-cos2=2
tanx=2+cos2
sinx=(2+cos2)cosx
sin^2x=(2+cos2)^2cos^2x
1-cos^2=(2+cos2)^2cosx^2
cos^2x=1/[(2+cos2)^2+1]
cosx=√1/[(2+cos2)^2+1]
sin^2x=(2+cos2)^2(1-sin^2x)
sin^2x[(2+cos2)^2+1]=(2+cos2)^2
sinx==(2+cos2)/√[(2+cos2)^2+1]
sinxcosx
=(2+cos2)/[(2+cos2)^2+1]

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解:太易了.因为由诱导公式有:
tan(π+x)+cos(3π+2)=tanx-cos2=2,所以tanx=2+cos2
再由倍角公式:
sinxcosx=1/2*sin(2x)=tanx/(1+tanx^2)
=(2+cos2)/[1+(2+cos2)^2]
=(2+cos2)/[5+4cos2+(cos2)^2]
=(2+cos2)/[5+4cos2+(1+cos4)/2]
=(4+2cos2)/(11+8cos2+cos4)

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