求解第3和第4题

1个回答

  • (3)

    因a与b均为非零实数,故

    (asinπ/5+bcosπ/5)/(acosπ/5-bsinπ/5) 两边均除以cosπ/5

    =(atanπ/5+b)/(a-btanπ/5) 两边均除以a

    =(b/a+tanπ/5)/(1-b/atanπ/5)

    ={tan[arctan(b/a)]+tanπ/5}/{1-tan[arctan(b/a)]*tanπ/5}

    =tan[arctan(b/a)+tanπ/5]=tan(8π/15)

    arctan(b/a)+π/5=8π/15+kπ

    arctan(b/a)=π/3+kπ

    tan[arctan(b/a)]=tan(π/3+kπ)

    b/a=√3

    (4)

    f(x)=sin平方x+acosx-(12a)-32=1-cosx平方+acosx-(12a)-32

    =-cosx平方+acosx-(12a)-12

    最大值为a平方的四分之一-(12a)-12=1

    a=1正负根号7