(sinx+3cosx)/(3cosx-sinx)=5,则sin^2x-sinxcosx的值是

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(sinx+3cosx)/(3cosx-sinx)=5,则sin^2x-sinxcosx的值是

(sinx+3cosx)/(3cosx-sinx)=5,则sin^2x-sinxcosx的值是
(sinx+3cosx)/(3cosx-sinx)=5,则sin^2x-sinxcosx的值是

(sinx+3cosx)/(3cosx-sinx)=5,则sin^2x-sinxcosx的值是
齐次式求值问题.
∵(sinx+3cosx)/(3cosx-sinx)=5
∴(tanx+3)/(3-tanx)=5 解得:tanx=2
∴sin²x-sinxcosx
=(sin²x-sinxcosx)/1
=(sin²x-sinxcosx)/(sin²x+cos²x)
=(tan²x-tanx)/(tan²x+1)
=(4-2)/(4+1)
=2/5

1

因为(sinx+3cosx)/(3cosx-sinx)=5,去分母,得:sinx+3cosx=15cosx-5sinx
化简,6sinx=12cosx,即:sinx=2cosx,两边同时除以cosx,得:tanx=2
所以:sin^2x-sinxcosx=(sin^2x-sinxcosx)/(sin^2x+cos^2x) (分子分母同时除以cos^2x)
=(tan^2x-tanx)/(tan^2x+1)
=(4-2)/(4+1)
=2/5