(sinx+cosx)/（sinx-cosx）=3 求sin^2(x)+sinxcosx
问题描述：
(sinx+cosx)/（sinx-cosx）=3 求sin^2(x)+sinxcosx
答
(sinx+cosx)/（sinx-cosx）=3在分子分母同除以sinx
解得1+tanx/1-tanx=3解得tanx=1/2
sinx^2+sinxcosx
=sinx^2+sinxcosx/sinx^2+cosx^2(在分子分母除以sinx^2)
=1+tanx/1+tanx^2(tanx=1/2)
=6/5
答
由已知条件，得出sinx=2cosx，令所求sin^2(x)+sinxcosx=A
将两端同除以sin^2x,得出1+cosx/sinx=A/sin^2x推出sin^2x=(2/3)A
将两端同除以cos^2x,得出4+sinx/cosx=A/cos^2x推出cos^2x=(1/6)A
将上两式相加，利用sin^2x+cos^2x=1，推出A=6/5
答
sinx+cosx=3sinx-3cosx
2sinx=4cosx
tanx=2
cosx^2=1/(1+tanx^2)=1/5
sinx^2=1-cosx^2=4/5
sinxcosx=tanxcosx^2=2/5
sinx^2+sinxcosx=4/5+2/5=6/5