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  • 1 # 83823堃

    解:原函數為:(1/2)[lnsin(x+π/4)+x]+C

    我只能輸入沒幾個字,要點是:cosx+sinx=2^(1/2)sin(x+π/4)

    令t=x+π/4 代入:dx=dt 就能得到結果

    分子分母同時除以cosx 得 1/(1+tanx)

    (-cosx+C)'=sinx,(sinx+C)'=cosx,

    sinx原函數是-cosx+C;

    cosx原函數是sinx+C.

    cosx的原函數是sinx, sinx的原函數是-cosx

  • 2 # y吃橘子嗎

    函數y=sinx-cosx+sinxcosx的值域為[-(1+2√2)/2,1]。

    解答過程如下:

    y=sinx-cosx+sinxcosx

    =√2[(√2/2)sinx-(√2/2)cosx]+½sin2x

    =√2sin(x-π/4)+½cos(π/2 -2x)

    =√2sin(x-π/4)+½cos[2(x-π/4)]

    =√2sin(x-π/4)+½[1-2sin²(x-π/4)]

    =-sin²(x-π/4)+√2sin(x-π/4)+½

    =-sin²(x-π/4)+√2sin(x-π/4)-½+1

    =-[sin(x-π/4) - √2/2]²+1

    sin(x-π/4)=√2/2時,y取得最大值,ymax=1

    sin(x-π/4)=-1時,y取得最小值,ymin=-(1+2√2)/2

    函數的值域為[-(1+2√2)/2,1]

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