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  • 三角函數(shù)題cot.

    三角函數(shù)題cot.
    已知A、B、C是三角形的三個(gè)內(nèi)角
    求證cot(A/2)+cot(B/2)+cot(C/2)=cot(A/2)*cot(B/2)*cot(C/2)
    數(shù)學(xué)人氣:706 ℃時(shí)間:2020-02-06 02:24:08
    優(yōu)質(zhì)解答
    cot(A/2)+cot(B/2)+cot(C/2)=cos(A/2)/sin(A/2)+cos(B/2)/sin(B/2)+cos(C/2)/sin(C/2)=[sin(B/2)cos(A/2)+cos(B/2)sin(A/2)]/sin(A/2)sin(B/2)+cos(C/2)/sin(C/2)=sin(A/2+B/2)/sin(A/2)sin(B/2)+cos(C/2)/sin(C/2)=sin(π/2-C/2)/sin(A/2)sin(B/2)+cos(C/2)/sin(C/2)=cos(C/2)/sin(A/2)sin(B/2)+cos(C/2)/sin(C/2)=[cos(C/2)sin(C/2)+sin(A/2)sin(B/2)cos(C/2)]/sin(A/2)sin(B/2)sin(C/2)={cos(C/2)[sin(c/2)+sin(A/2)sin(B/2)]
    }/sin(A/2)sin(B/2)sin(C/2)={cos(C/2)[cos(B/2)cos(A/2)-sin(A/2)sin(B/2)+sin(A/2)sin(B/2)]}/sin(A/2)sin(B/2)sin(C/2)=cos(A/2)cos(B/2)cos(C/2)/sin(A/2)sin(B/2)sin(C/2)=cot(A/2)*cot(B/2)*cot(C/2)
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