1 prove the following identities:

1个回答

  • 1.

    a.

    cosh(2x)

    = [e^(2x) + e^(-2x)]/2

    = [(e^x)²+(e^-x)²]/2

    = [(e^x + e^-x)² + (e^x - e^-x)²]/4

    = (e^x + e^-x)²/4 + (e^x - e^-x)²/4

    = cosh²(x) + sinh²(x)

    b.

    cosh(x+y)

    = [e^(x+y) + e^(-x-y)]/2

    = [e^x * e^y + e^-x * e^-y]/2

    = [e^x * e^y + e^x * e^-y +e^-x * e^y + e^-x * e^-y]/4 + [e^x * e^y - e^x * e^-y -e^-x * e^y + e^-x * e^-y]/4

    = (e^x + e^-x)(e^y + e^-y)/4 + (e^x - e^-x)(e^y - e^-y)/4

    = cosh(x)cosh(y)+sinh(x)sinh(y)

    2.

    cosh(x) = (e^x + e^-x)/2

    令y = cosh(x),即x = cosh^-1(y),则2y = e^x + e^-x

    (e^x)² - 2ye^x + 1 = 0

    e^x = y ± √(y²-1)

    x = ln[y ± √(y²-1)] = cosh^-1(y)

    即cosh^-1(x) = ln[x ± √(x²-1)]

    因为x - √(x²-1) = 1/[x + √(x²-1)] 恒不大于1(小于或等于1)

    则ln[x - √(x²-1)] ≤ 0

    一般取该函数的正支,即cosh^-1(x) = ln[x + √(x²-1)]

    3.

    令y = sech^-1(x),则x = sech(y),dx = [sech(y)]'dy

    而d[sech^-1(x)]/dx = dy/dx = 1/[sech(y)]'

    sech(y) = 2/(e^y + e^-y)

    [sech(y)]' = [2/(e^y + e^-y)]' = -2(e^y - e^-y)/(e^y + e^-y)² = -sinh(y)/cosh²(y)

    dy/dx = 1/[sech(y)]' = -cosh²(y)/sinh(y)

    因为x = sech(y),所以cosh(y) = 1/x,sinh(y) = √(1 - 1/x²)

    代入上式得到dy/dx = -1/x² * 1/√(1 - 1/x²) = -1/[x√(x²-1)]

    即d[sech^-1(x)]/dx = -1/[x√(x²-1)]

    4.

    ∫(-3,3)e^xt dt

    = 1/x * ∫(-3,3)e^xt dxt

    = 1/x * e^xt|(-3,3)

    = 1/x * (e^3x - e^-3x)

    = 2sinh(3x)/x