求tanπ/8 + cotπ/12的值

2个回答

  • tanπ/8 + cotπ/12

    =tan(1/2*π/4)+cot(π/3-π/4)

    =√[(1-cosπ/4)/(1+cosπ/4)]+(cotπ/3*cotπ/4+1)/(cotπ/4-cotπ/3)

    =√[(1-√2/2)/(1+√2/2)]+(√3/3*1+1)/(1-√3/3)

    =√3+√2+1

    三角函数公式

    两角和公式

    tan(A+B)=(tanA+tanB)/(1-tanAtanB)

    tan(A-B)=(tanA-tanB)/(1+tanAtanB)

    ctg(A+B)=(ctgActgB-1)/(ctgB+ctgA)

    ctg(A-B)=(ctgActgB+1)/(ctgB-ctgA)

    半角公式

    tan(A/2)=√((1-cosA)/((1+cosA))

    tan(A/2)=-√((1-cosA)/((1+cosA))

    ctg(A/2)=√((1+cosA)/((1-cosA))

    ctg(A/2)=-√((1+cosA)/((1-cosA))