高一数学单调区间请详细解答,谢谢! (1 17:3:55)

显示全部楼层 · 2009-8-1 17:35:17

求函数y=sinxcosx÷(1+sinx-cosx)的单调区间

千问 · 2009-8-1 17:35:17

y=sinxcosx/(1+sinx-cosx) =2sin(x/2)cos(x/2)cosx/(2sin^2(x/2)+2sin(x/2)cos(x/2)) =cos(x/2)cosx/(sin(x/2)+cos(x/2)) =cos(x/2){cos^2(x/2)-sin^2(x/2)}/(sin(x/2) +cos(x/2)) =cos(x/2)[cos(x/2)-sin(x/2)] =(1+cosx)/2-sinx/2 =√2 cos(x+π/4)/2+1/2 画图知：（答题时这句话可省掉）y递增时，π+2kπ≤x+π/4≤2π+2kπ所以3π/4≤x≤7π/4+2kπy递减时2kπ≤x+π/4≤π+...

千问 · 2009-8-1 17:35:17

sinxcosx/(1+sinx-cosx)=2sin(x/2)cos(x/2)cosx/(2sin^2(x/2)+2sin(x/2)cos(x/2))=cos(x/2)cosx/(sin(x/2)+cos(x/2))=cos(x/2){cos^2(x/2)-sin^2(x/2)}/(sin(x/2)+cos(x/2))=cos(x/2)[c...

千问 · 2009-8-1 17:35:17

解答:换元法)令t=sinx+cosx,则t2=(sinx+cosx)2=1+2sinxcosx=1+sin2x≤2, ∴-√2≤t≤√2,由于即∴sinxcosx=(t2-1)/2,∵y=( sinxcosx) /(sinx+cosx-1)=(t2-1)/2(t-1)=( t+1)/2, -√2≤t≤...

千问 · 2009-8-1 17:35:17

应该先化简，可以化成tan（x/2）*cosx，再根据同增异减的原则，在区间[2kπ，π +2kπ]为减区间...

高一 数学 单调区间 请详细解答,谢谢! (1 17:3:55)

高一数学单调区间请详细解答,谢谢! (1 17:3:55)